Maharashtra Board Class 11 Physics Chapter 9 Optics Solutions

1. Multiple Choise Questions

Question 1. As per recent understanding light consists of
(A) rays
(B) waves
(C) corpuscles
(D) photons obeying the rules of waves
Answer: (D) photons obeying the rules of waves
In simple words: Modern physics describes light as having a dual nature, behaving both as particles (photons) and waves simultaneously.

🎯 Exam Tip: Understanding the dual nature of light is fundamental, often tested in basic concept questions.

Question 2. Consider the optically denser lenses P, Q, R and S drawn below. According to Cartesian sign convention which of these have positive focal length?
ℹ️ चित्र व्याख्या (Diagram Explanation): यहाँ चार लेंसों, P, Q, R और S, को दर्शाया गया है। लेंस P एक द्विउत्तल लेंस है, Q एक समतल-उत्तल लेंस है, R एक समतल-अवतल लेंस है, और S एक द्विअवतल लेंस है। इन लेंसों की आकृतियाँ उनकी प्रकाशीय सघनता को दर्शाती हैं और प्रश्न में पूछा गया है कि इनमें से किस लेंस की फोकस दूरी कार्टेशियन चिह्न परिपाटी के अनुसार धनात्मक होगी।
(A) OnlyP
(B) Only P and Q
(C) Only P and R
(D) Only Q and S
Answer: (B) Only P and Q
In simple words: Convex lenses (P and Q), which are thicker in the middle, converge light and therefore have a positive focal length according to the Cartesian sign convention.

🎯 Exam Tip: Remember that converging lenses (convex lenses) have positive focal lengths, while diverging lenses (concave lenses) have negative focal lengths in the Cartesian sign convention.

Question 3. Two plane mirrors are inclined at angle 40° between them. Number of images seen of a tiny object kept between them is
(A) Only 8
(B) Only 9
(C) 8 or 9
(D) 9 or 10
Answer: (C) 8 or 9
In simple words: The number of images formed by two inclined plane mirrors depends on the angle between them and whether the object is placed on the angle bisector. If \( \frac{360}{\theta} \) is an even integer, the number of images is \( \frac{360}{\theta} - 1 \). If it's an odd integer, it's \( \frac{360}{\theta} \) if off the bisector, or \( \frac{360}{\theta} - 1 \) if on the bisector. For 40°, \( \frac{360}{40} = 9 \) (odd), so it can be 8 or 9.

🎯 Exam Tip: For problems involving images formed by inclined mirrors, always calculate \( n = \frac{360}{\theta} \). If n is even, images = \( n-1 \). If n is odd, images = \( n \) (object off bisector) or \( n-1 \) (object on bisector).

Question 4. A concave mirror of curvature 40 cm, used for shaving purpose produces image of double size as that of the object. Object distance must be
(A) 10 cm only
(B) 20 cm only
(C) 30 cm only
(D) 10 cm or 30 cm
Answer: (D) 10 cm or 30 cm
In simple words: For a concave mirror, to produce an image of double size, the object can be placed between the pole and focal point (virtual, magnified image) or between the focal point and center of curvature (real, magnified image). The focal length is \( f = R/2 = 40/2 = 20 \) cm. Using magnification \( m = -\frac{v}{u} = 2 \) (virtual) or \( m = -\frac{v}{u} = -2 \) (real), and the mirror formula \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \), we can solve for u.

🎯 Exam Tip: Remember that for concave mirrors, magnification can be positive (virtual image) or negative (real image). For double size, \( m = +2 \) or \( m = -2 \), leading to two possible object positions.

Question 5. Which of the following aberrations will NOT occur for spherical mirrors?
(A) Chromatic aberration
(B) Coma
(C) Distortion
(D) Spherical aberration
Answer: (A) Chromatic aberration
In simple words: Chromatic aberration occurs due to the dispersion of light (different colors having different refractive indices) when passing through a lens. Since mirrors reflect light and do not rely on refraction, they do not suffer from chromatic aberration.

🎯 Exam Tip: Chromatic aberration is a defect specific to lenses because it depends on the wavelength-dependent refractive index of the lens material. Mirrors do not exhibit this aberration.

Question 6. There are different fish, monkeys and water of the habitable planet of the star Proxima b. A fish swimming underwater feels that there is a monkey at 2.5 m on the top of a tree. The same monkey feels that the fish is 1.6 m below the water surface. Interestingly, height of the tree and the depth at which the fish is swimming are exactly same. Refractive index of that water must be
(A) 6/5
(B) 5/4
(C) 4/3
(D) 7/5
Answer: (B) 5/4
In simple words: This problem involves apparent depth and height due to refraction. The fish sees the monkey closer than it actually is (apparent height from water is less than real height), and the monkey sees the fish shallower than it actually is (apparent depth from air is less than real depth). Using the given heights and depths, and the formula for apparent depth/height in relation to the refractive index, the refractive index of water can be calculated as 5/4.

🎯 Exam Tip: For apparent depth/height problems, remember the formula \( \mu = \frac{\text{real depth}}{\text{apparent depth}} \) or \( \mu = \frac{\text{apparent height}}{\text{real height}} \). Consistency in which medium is taken as '1' and '2' is crucial for accurate calculations.

Question 7. Consider following phenomena/applications: P) Mirage, Q) rainbow, R) Optical fibre and S) glittering of a diamond. Total internal reflection is involved in
(A) Only R and S
(B) Only R
(C) Only P, R and S
(D) all the four
Answer: (A) Only R and S
In simple words: Total internal reflection (TIR) is the phenomenon where light traveling from a denser to a rarer medium is reflected back into the denser medium if the angle of incidence exceeds the critical angle. Optical fibres work entirely on TIR to transmit light, and the glittering of a diamond is due to multiple TIRs within its facets. Mirage involves refraction, and a rainbow involves both refraction and reflection.

🎯 Exam Tip: Clearly distinguish between phenomena primarily caused by refraction (like mirage) and those heavily dependent on total internal reflection (like optical fibers and diamond brilliance). Rainbows involve both refraction and reflection.

Question 8. A student uses spectacles of number -2 for seeing distant objects. Commonly used lenses for her/his spectacles are
(A) bi-concave
(B) piano concave
(C) concavo-convex
(D) convexo-concave
Answer: (A) bi-concave
In simple words: A spectacle power of -2 indicates a diverging lens, which is used to correct myopia (nearsightedness), where distant objects appear blurry. Bi-concave lenses are diverging lenses, meaning they are thicker at the edges and thinner in the middle, and are typically used for this purpose.

🎯 Exam Tip: Negative power spectacles indicate a diverging lens, which is used to correct myopia. Such lenses are typically biconcave or plano-concave.

Question 9. A spherical marble, of refractive index 1.5 and curvature 1.5 cm, contains a tiny air bubble at its centre. Where will it appear when seen from outside?
(A) 1 cm inside
(B) at the centre
(C) 5/3 cm inside
(D) 2 cm inside
Answer: (B) at the centre
In simple words: When an object is at the center of curvature of a spherical refracting surface, its image also forms at the center of curvature. An air bubble at the center of a spherical marble acts as an object at the center of curvature when viewed from outside, so it appears to be at the center.

🎯 Exam Tip: For objects placed at the center of curvature of a spherical refracting surface, the image is formed at the same location, irrespective of the refractive indices, as long as light passes through it without deviation at that point.

Question 10. Select the WRONG statement.
(A) Smaller angle of prism is recommended for greater angular dispersion.
(B) Right angled isosceles glass prism is commonly used for total internal reflection.
(C) Angle of deviation is practically constant for thin prisms.
(D) For emergent ray to be possible from the second refracting surface, certain minimum angle of incidence is necessary from the first surface.
Answer: (A) Smaller angle of prism is recommended for greater angular dispersion.
In simple words: Angular dispersion is generally proportional to the angle of the prism. Therefore, a *larger* angle of prism, not a smaller one, would typically lead to greater angular dispersion. The other statements are generally true regarding prisms and total internal reflection.

🎯 Exam Tip: Angular dispersion is typically proportional to the prism angle (A) and the difference in refractive indices for extreme colors (\( n_v - n_r \)). A smaller prism angle will result in *less* angular dispersion.

Question 11. Angles of deviation for extreme colours are given for different prisms. Select the one having maximum dispersive power of its material.
(A) 7°, 10°
(B) 8°, 11°
(C) 12°, 16°
(D) 10°, 14°
Answer: (A) 7°, 10°
In simple words: Dispersive power (\( \omega \)) is defined as the ratio of angular dispersion to the mean deviation. Angular dispersion is the difference between the deviation of violet and red light (\( \delta_v - \delta_r \)). Mean deviation is approximately the average of the extreme deviations (\( \frac{\delta_v + \delta_r}{2} \)). To find maximum dispersive power, we need to calculate \( \omega \) for each option.

🎯 Exam Tip: Dispersive power is \( \omega = \frac{\delta_v - \delta_r}{\delta_{mean}} \), where \( \delta_{mean} = \frac{\delta_v + \delta_r}{2} \). Calculate this ratio for each option to find the maximum value.

Question 12. Which of the following is not involved in formation of a rainbow?
(A) refraction
(B) angular dispersion
(C) angular deviation
(D) total internal reflection
Answer: (D) total internal reflection
In simple words: Rainbow formation involves refraction, dispersion, and *internal reflection* (not necessarily total internal reflection) of sunlight within water droplets. While internal reflection occurs, it's not strictly total internal reflection that is essential, rather a single internal reflection.

🎯 Exam Tip: Rainbows are formed by a combination of refraction, dispersion, and a single *internal reflection* inside water droplets. While total internal reflection can occur, it is not a prerequisite for rainbow formation; simple internal reflection is sufficient.

Question 13. Consider following statements regarding a simple microscope:
(P) It allows us to keep the object within the least distance of distinct vision.
(Q) Image appears to be biggest if the object is at the focus.
(R) It is simply a convex lens.
(A) Only (P) is correct
(B) Only (P) and (Q) are correct
(C) Only (Q) and (R) are correct
(D) Only (P) and (R) are correct
Answer: (D) Only (P) and (R) are correct
In simple words: A simple microscope is essentially a convex lens. It works by allowing the object to be placed closer to the eye than the least distance of distinct vision, thus creating a magnified virtual image. The image appears biggest when the image is formed at the least distance of distinct vision, which typically happens when the object is placed just inside the focal point.

🎯 Exam Tip: A simple microscope uses a convex lens. For maximum magnification, the image should be formed at the least distance of distinct vision (D), which occurs when the object is slightly inside the focal length.

2. Answer the following questions.

Question 1. As per recent development, what is the nature of light? Wave optics and particle nature of light are used to explain which phenomena of light respectively?
Answer:
1. As per recent development, it is now an established fact that light possesses dual nature. Light consists of energy carrier photons. These photons follow the rules of electromagnetic waves.
2. Wave optics explains the phenomena of light such as, interference, diffraction, polarisation, Doppler effect etc.
3. Particle nature of light can be used to explain phenomena like photoelectric effect, emission of spectral lines, Compton effect etc.
In simple words: Light has a dual nature, behaving as both a wave (explained by wave optics for phenomena like interference and diffraction) and a particle (photons, explained by particle nature for phenomena like the photoelectric effect).

🎯 Exam Tip: Be precise in associating specific light phenomena with either its wave nature (e.g., interference, diffraction) or particle nature (e.g., photoelectric effect, Compton effect).

Question 2. Which phenomena can be satisfactorily explained using ray optics?
Answer:
Ray optics or geometrical optics: Ray optics can be used for understanding phenomena like reflection, refraction, double refraction, total internal reflection, etc.
In simple words: Ray optics, also known as geometrical optics, is used to explain light phenomena such as reflection, refraction, and total internal reflection by treating light as rays traveling in straight lines.

🎯 Exam Tip: Ray optics is applicable when the wavelength of light is much smaller than the dimensions of the optical instruments, making it suitable for macroscopic phenomena like reflection and simple refraction.

Question 3. What is focal power of a spherical mirror or a lens? What may be the reason for using \( P = \frac{1}{f} \) its expression?
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यहाँ दो अलग-अलग आकृतियाँ दी गई हैं जो लेंस और दर्पण के प्रकाशीय व्यवहार को दर्शाती हैं। पहली आकृति में, एक उत्तल लेंस प्रकाश किरणों को फोकस बिंदु F पर अभिसरित करता हुआ दिखाया गया है, जबकि दूसरी आकृति में, एक अवतल लेंस प्रकाश किरणों को अपसरित करता हुआ दिखाया गया है। यह दर्शाता है कि लेंस और दर्पण कैसे प्रकाश को मोड़ते हैं और उनकी फोकल पावर क्या होती है।

1. Converging or diverging ability of a lens or of a mirror is defined as its focal power.
2. This implies, more the power of any spherical mirror or a lens, the more is its ability to converge or diverge the light that passes through it.
3. In case of convex lens or concave mirror, more the convergence, shorter is the focal length as shown in the figure.
4. Similarly, in case of concave lens or convex mirror, more the divergence, shorter is the focal length.
5. This explains that the focal power of any spherical lens or mirror is inversely proportional to the focal length.
6. Hence, the expression of focal power is given by the formula, \( P = \frac{1}{f} \).
In simple words: Focal power measures how strongly a lens or mirror converges or diverges light. It is defined as the reciprocal of the focal length ( \( P = \frac{1}{f} \) ), because a shorter focal length indicates a stronger ability to bend light.

🎯 Exam Tip: Focal power is a measure of how "strong" a lens or mirror is. Remember to use focal length in meters when calculating power in diopters.

Question 4. At which positions of the objects do spherical mirrors produce (i) diminished image (ii) magnified image?
Answer:
i. Amongst the two types of spherical mirrors, convex mirror always produces a diminished image at all positions of the object.
ii. Concave mirror produces diminished image when object is placed:
- Beyond radius of curvature (i.e., \( u > 2f \))
- At infinity (i.e., \( u = \infty \))
iii. Concave mirror produces magnified image when object is placed:
- between centre of curvature and focus (i.e., \( 2f > u > f \))
- between focus and pole of the mirror (i.e., \( u < f \))
In simple words: Convex mirrors always create diminished images. Concave mirrors create diminished images when the object is beyond the center of curvature or at infinity, and magnified images when the object is between the focal point and center of curvature or between the focal point and pole.

🎯 Exam Tip: Memorize the image formation rules for concave and convex mirrors, especially the conditions for real/virtual, inverted/erect, and diminished/magnified images based on object position.

Question 5. State the restrictions for having images produced by spherical mirrors to be appreciably clear.
Answer:
i. In order to obtain clear images, the formulae for image formation by mirrors or lens follow the given assumptions:
- Objects and images are situated close to the principal axis.
- Rays diverging from the objects are confined to a cone of very small angle.
- If there is a parallel beam of rays, it is paraxial, i.e., parallel and close to the principal axis.
ii. In case of spherical mirrors (excluding small aperture spherical mirrors), rays farther from the principle axis do not remain parallel to the principle axis. Thus, the third assumption is not followed and the focus gradually shifts towards the pole.
iii. The relation \( (f = \frac{R}{2}) \) giving a single point focus is not followed and the image does not get converged at a single point resulting into a distorted or defective image.
iv. This defect arises due to the spherical shape of the reflecting surface.
In simple words: Clear images from spherical mirrors require objects and images to be near the principal axis, and rays to be paraxial (close to and parallel to the principal axis). Deviations from these assumptions cause distortions like spherical aberration.

🎯 Exam Tip: The thin lens/mirror formulas are based on paraxial ray approximation. When these approximations break down, aberrations occur, leading to unclear images.

Question 6. Explain spherical aberration for spherical mirrors. How can it be minimized? Can it be eliminated by some curved mirrors?
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र गोलाकार दर्पणों में होने वाले गोलीय विपथन (spherical aberration) को दर्शाता है। इसमें दिखाया गया है कि कैसे दर्पण के केंद्र के पास (पैरैक्सियल किरणें) और किनारे के पास (मार्जिनल किरणें) से आने वाली समानांतर प्रकाश किरणें एक ही फोकस बिंदु पर अभिसरित नहीं होतीं। मार्जिनल किरणें ध्रुव के करीब फोकस होती हैं (FM) जबकि पैरैक्सियल किरणें ध्रुव से दूर फोकस होती हैं (Fp)। इस अंतर के कारण एक स्पष्ट बिंदु छवि के बजाय एक धुंधली छवि बनती है।

1. In case of spherical mirrors (excluding small aperture spherical mirror), The rays coming from a distant object farther from principal axis do not remain parallel to the axis. Thus, the focus gradually shifts towards the pole.
2. The assumption for clear image formation namely, 'If there is a parallel beam of rays, it is paraxial, i.e., parallel and close to the principal axis', is not followed in this case.
3. The relation \( f (f = \frac{R}{2}) \) giving a single point focus is not followed and the image does not get converged at a single point resulting into a distorted or defective image.
4. This phenomenon is known as spherical aberration.
5. It occurs due to spherical shape of the reflecting surface, hence known as spherical aberration.
6. The rays near the edge of the mirror converge at focal point \( F_M \) Whereas, the rays near the principal axis converge at point \( F_P \). The distance between \( F_M \) and \( F_P \) is measured as the longitudinal spherical aberration.
7. In spherical aberration, single point image is not possible at any point on the screen and the image formed is always a circle.
8. At a particular location of the screen (across AB in figure), the diameter of this circle is minimum. This is called the circle of least confusion. Radius of this circle is transverse spherical aberration.

Remedies for Spherical Aberration:

1. Spherical aberration can be minimized by reducing the aperture of the mirror.
2. Spherical aberration in curved mirrors can be completely eliminated by using a parabolic mirror.
In simple words: Spherical aberration is a defect where parallel rays striking a spherical mirror at different distances from the principal axis do not converge to a single focal point, leading to a blurred image. It can be minimized by reducing the mirror's aperture or entirely eliminated by using parabolic mirrors.

🎯 Exam Tip: Spherical aberration is inherent to spherical surfaces for wide beams. Parabolic mirrors are crucial for applications requiring a perfect point focus, as they are free from spherical aberration.

Question 7. Define absolute refractive index and relative refractive index. Explain in brief with an illustration for each.
Answer:
i. Absolute refractive index of a medium is defined as the ratio of speed of light in vacuum to that in the given medium.
ii. A stick or pencil kept obliquely in a glass containing water appears broken as its part in water appears to be raised.
iii. As the speed of light is different in two media, the rays of light coming from water undergo refraction at the boundary separating two media.
iv. Consider speed of light to be v in water and c in air. (Speed of light in air - speed of light in vacuum)
\( \therefore \) refractive index of water \( = \frac{N_w}{n_s} = \frac{N_w}{N_{vacuum}} = \frac{c}{v} \)
v. Relative refractive index of a medium 2 is the refractive index of medium 2 with respect to medium 1 and it is defined as the ratio of speed of light \( v_1 \) in medium 1 to its speed \( v_1 \) in medium 2.
\( \therefore \) Relative refractive index of medium 2,
\( ^1 n_2 = \frac{v_1}{v_2} \)
vi. Consider a beaker filled with water of absolute refractive index \( n_1 \) kept on a transparent glass slab of absolute refractive index \( n_2 \).
vii. Thus, the refractive index of water with respect to that of glass will be,
\( ^W n_2 = \frac{n_2}{n_1} = \frac{c/v_2}{c/v_1} = \frac{v_1}{v_2} \)
In simple words: Absolute refractive index is the ratio of light speed in vacuum to its speed in a medium. Relative refractive index is the ratio of light speeds in two different media, showing how much light bends when passing from one to another.

🎯 Exam Tip: Clearly state the definitions and use the correct ratio of speeds for both absolute (\( \mu = c/v \)) and relative (\( \mu_{21} = v_1/v_2 \)) refractive indices. Illustrations like a bent pencil are helpful for explaining refraction.

Question 8. Explain 'mirage' as an illustration of refraction.
Answer:
i. On a hot clear Sunny day, along a level road, there appears a pond of water ahead of the road. However, if we physically reach the spot, there is nothing but the dry road and water pond again appears some distance ahead. This illusion is called mirage.
ii. Mirage results from the refraction of light through a non-uniform medium.
iii. On a hot day the air in contact with the road is hottest and as we go up, it gets gradually cooler. The refractive index of air thus decreases with height. Hot air tends to be less optically dense than cooler air which results into a non-uniform medium.
iv. Light travels in a straight line through a uniform medium but refracts when traveling through a non-uniform medium.
v. Thus, the ray of light coming from the top of an object get refracted while travelling downwards into less optically dense air and become more and more horizontal as shown in Figure.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र 'मरीचिका' की घटना को दर्शाता है। इसमें दिखाया गया है कि एक गर्म दिन में, जमीन के पास की हवा ऊपर की हवा की तुलना में गर्म और विरल होती है, जिससे प्रकाश का अपवर्तनांक कम हो जाता है। एक दूर के पेड़ से आने वाली प्रकाश किरणें धीरे-धीरे ऊपर की ओर मुड़ती हैं क्योंकि वे अलग-अलग घनत्व वाली हवा की परतों से गुजरती हैं। एक पर्यवेक्षक को यह आभास होता है कि प्रकाश जमीन से परावर्तित हो रहा है, जिससे पानी के पूल का भ्रम पैदा होता है।

vi. As it almost touches the road, it bends (refracts) upward. Then onwards, upward bending continues due to denser air.
vii. As a result, for an observer, it appears to be coming from below thereby giving an illusion of reflection from an (imaginary) water surface.
In simple words: Mirage is an optical illusion where distant objects appear inverted or a pool of water appears on a hot road. It occurs due to the refraction of light through layers of air with varying refractive indices, caused by temperature differences, bending light rays to the observer's eye from unexpected directions.

🎯 Exam Tip: Emphasize that mirage is a phenomenon of refraction, not reflection, and occurs due to the gradual change in the refractive index of air with temperature gradients.

Question 9. Under what conditions is total internal reflection possible? Explain it with a suitable example. Define critical angle of incidence and obtain an expression for it.
Answer:
Conditions for total internal reflection:
i. The light ray must travel from denser medium to rarer medium.
ii. The angle of incidence in the denser medium must be greater than critical angle for the given pair of media.

Total internal reflection in optical fibre:
iii. Consider an optical fibre made up of core of refractive index \( n_1 \) and cladding of refractive index \( n_2 \) such that, \( n_1 > n_2 \).
iv. When a ray of light is incident from a core (denser medium), the refracted ray is bent away from the normal.
v. At a particular angle of incidence \( i_c \) in the denser medium, the corresponding angle of refraction in the rarer medium is 90°.
vi. For angles of incidence greater than \( i_c \), the angle of refraction become larger than 90° and the ray does not enter into rarer medium at all but is reflected totally into the denser medium as shown in figure.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र पूर्ण आंतरिक परावर्तन (Total Internal Reflection) की अवधारणा को दर्शाता है। इसमें एक प्रकाश किरण को एक सघन माध्यम (जैसे कोर) से एक विरल माध्यम (जैसे क्लैडिंग) की ओर यात्रा करते हुए दिखाया गया है। जैसे-जैसे आपतन कोण बढ़ता है, अपवर्तन कोण भी बढ़ता है, और एक महत्वपूर्ण कोण (\( i_c \)) पर, अपवर्तित किरण सतह के समानांतर (90°) हो जाती है। यदि आपतन कोण \( i_c \) से अधिक हो जाए, तो किरण पूरी तरह से सघन माध्यम में परावर्तित हो जाती है, जिसे पूर्ण आंतरिक परावर्तन कहते हैं।

critical angle of incidence and obtain an expression:
i. Critical angle for a pair of refracting media can be defined as that angle of incidence in the denser medium for which the angle of refraction in the rarer medium is 90°.
ii. Let n be the relative refractive index of denser medium with respect to the rarer.
iii. Then, according to Snell's law,
\( n = \frac{n_{denser}}{n_{rarer}} = \frac{\sin r}{\sin i_c} = \frac{\sin 90°}{\sin i_c} \)
\( \therefore \sin (i_c) = \frac{1}{n} \)
In simple words: Total internal reflection occurs when light travels from a denser to a rarer medium, and the angle of incidence exceeds the critical angle. The critical angle is the angle of incidence in the denser medium where the angle of refraction in the rarer medium is 90°, given by \( \sin(i_c) = 1/n \).

🎯 Exam Tip: Clearly state the two conditions for TIR and derive the critical angle expression using Snell's Law. Optical fibers are a classic example of TIR in action.

Question 10. Describe construction and working of an optical fibre. What are the advantages of optical fibre communication over electronic communication?
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक ऑप्टिकल फाइबर की संरचना को दर्शाता है, जिसमें एक अत्यंत पतला, पारदर्शी और लचीला 'कोर' होता है, जो एक प्रकाशीय रूप से विरल लचीले कवर जिसे 'क्लैडिंग' कहते हैं, से घिरा होता है। कोर, क्लैडिंग से सघन होता है। फाइबर की कुल मोटाई लगभग 1 मिमी होती है और यह कई किलोमीटर तक लंबा हो सकता है, जो संकेत संचरण के लिए उपयोग किया जाता है।

Construction:

1. An optical fibre consists of an extremely thin, transparent and flexible core surrounded by an optically rarer flexible cover called cladding.
2. For protection, the whole system is coated by a buffer and a jacket.
3. Entire thickness of the fibre is less than half a mm.
4. Many such fibres can be packed together in an outer cover.

Working:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र ऑप्टिकल फाइबर के काम करने के सिद्धांत, पूर्ण आंतरिक परावर्तन (Total Internal Reflection) को दर्शाता है। एक ऑप्टिकल सिग्नल (प्रकाश की किरण) कोर में प्रवेश करती है और क्लैडिंग के साथ कोर की आंतरिक सतह पर बार-बार पूर्ण आंतरिक परावर्तन का सामना करती है। इन परावर्तनों के कारण, प्रकाश सिग्नल कई किलोमीटर की दूरी तक बिना किसी महत्वपूर्ण नुकसान के फाइबर के माध्यम से यात्रा करता है।

1. Working of optical fibre is based on the principle of total internal reflection.
2. An optical signal (a ray of light) entering the core suffers multiple total internal reflections before emerging out after a several kilometres.
3. The optical signal travels with the highest possible speed in the material.
4. The emerged optical signal has extremely low loss in signal strength.

Advantages of optical fibre communication over electronic communication:

1. Broad bandwidth (frequency range): For TV signals, a single optical fibre can carry over 90000 independent signals (channels).
2. Immune to EM interference: Optical fibre being electrically non-conductive, does not pick up nearby EM signals.
3. Low attenuation loss: loss being lower than 0.2 dB/km, a single long cable can be used for several kilometres.
4. Electrical insulator: Optical fibres being electrical insulators, ground loops of metal wires or lightning do not cause any harm.
5. Theft prevention: Optical fibres do not use copper or other expensive material which are prone to be robbed.
6. Security of information: Internal damage is most unlikely to occur, keeping the information secure.
In simple words: Optical fibers consist of a core and cladding (denser core, rarer cladding) and work on the principle of total internal reflection, guiding light signals over long distances with minimal loss. They offer advantages like broad bandwidth, immunity to electromagnetic interference, low signal loss, and enhanced security over electronic communication.

🎯 Exam Tip: When describing optical fibers, clearly explain both the construction (core, cladding, buffer) and the working principle (total internal reflection). Highlight key advantages such as bandwidth, signal integrity, and immunity to EMI.

Question 11. Why are prism binoculars preferred over traditional binoculars? Describe its working in brief.
Answer:
1. Traditional binoculars use only two cylinders. Distance between the two cylinders can't be greater than that between the two eyes. This creates a limitation of field of view.
2. A prism binocular has two right angled glass prisms which apply the principle of total internal reflection.
3. The incident light rays are reflected internally twice giving the viewer a wider field of view. For this reason, prism binoculars are preferred over traditional binoculars.

Working:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक प्रिज्म दूरबीन की कार्यप्रणाली को दर्शाता है। इसमें चार समद्विबाहु, समकोण प्रिज्म (1, 2, 3, 4) का उपयोग किया जाता है। प्रकाश किरणें (Wider input range) पहले प्रिज्म 1 और 4 से पूर्ण आंतरिक रूप से परावर्तित होती हैं, फिर प्रिज्म 2 और 3 से पुनः पूर्ण आंतरिक परावर्तन होता है, जिससे अंतिम छवि बनती है। यह व्यवस्था आँखों के बीच की औसत दूरी से अधिक चौड़ा इनपुट रेंज प्रदान करती है, जिससे देखने का क्षेत्र बढ़ जाता है।

1. The prism binoculars consist of 4 isosceles, right angled prisms of material having critical angle less than 45°.
2. The prism binoculars have a wider input range compared to traditional binoculars.
3. The light rays incident on the prism binoculars, first get total internally reflected by the isosceles, right angled prisms 1 and 4.
4. These reflected rays undergo another total internal reflection by prisms 2 and 3 to form the final image.
In simple words: Prism binoculars are preferred because they use prisms to reflect light multiple times internally, effectively increasing the path length and thus the separation between objective lenses, which provides a wider field of view compared to traditional binoculars.

🎯 Exam Tip: The key advantage of prism binoculars is their ability to increase the effective baseline (distance between objectives) using total internal reflection, leading to a wider field of view and more compact design.

Question 12. A spherical surface separates two transparent media. Derive an expression that relates object and image distances with the radius of curvature for a point object. Clearly state the assumptions, if any.
Answer:
i. Consider a spherical surface YPY' of radius curvature R, separating two transparent media of refractive indices \( n_1 \) and \( n_2 \) respectively with \( n_1 < n_2 \).
ii. P is the pole and X'PX is the principal axis. A point object O is at a distance u from the pole, in the medium of refractive index \( n_1 \).
iii. In order to minimize spherical aberration, we consider two paraxial rays.
iv. The ray OP along the principal axis travels undeviated along PX. Another ray OA strikes the surface at A.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक गोलाकार अपवर्तक सतह को दर्शाता है जो दो पारदर्शी माध्यमों \( n_1 \) और \( n_2 \) को अलग करती है, जहाँ \( n_1 < n_2 \)। एक बिंदु वस्तु O से आने वाली प्रकाश किरणें सतह पर अपवर्तित होती हैं और एक वास्तविक छवि I बनाती हैं। चित्र में आपतन कोण (\( \alpha \)), परावर्तित कोण (\( \beta \)) और अपवर्तित कोण (\( \gamma \)) के साथ-साथ वक्रता त्रिज्या R और वस्तु व छवि दूरी u और v को भी दिखाया गया है। यह आरेख एक गोलाकार सतह से अपवर्तन के लिए सूत्र की व्युत्पत्ति को समझने में मदद करता है।

v. As \( n_1 < n_2 \), the ray deviates towards the normal (CAN), travels along AZ and real image of point object O is formed at I.
vi. Let \( \alpha \), \( \beta \) and \( \gamma \) be the angles subtended by incident ray, normal and refracted ray with the principal axis.
\( \therefore i = (\alpha + \beta) \) and \( r = (\beta - \gamma) \)
vii. As, the rays are paraxial, all the angles can be considered to be very small.
i.e., \( \sin i \approx i \) and \( \sin r \approx r \)
Angles \( \alpha \), \( \beta \) and \( \gamma \) can also be expressed as,
\( \alpha = \frac{\text{arc PA}}{\text{OP}} = \frac{\text{arc PA}}{u} \)
\( \beta = \frac{\text{arc PA}}{\text{PC}} = \frac{\text{arc PA}}{R} \)
and \( \gamma = \frac{\text{arc PA}}{\text{PI}} = \frac{\text{arc PA}}{v} \)
viii. According to Snell's law,
\( n_1 \sin (i) = n_2 \sin (r) \)
For small angles, Snell's law can be written
as, \( n_1 i = n_2 r \)
\( \therefore n_1 (\alpha + \beta) = n_2 (\beta - \gamma) \)
\( \therefore (n_2 - n_1)\beta = n_1 \alpha + n_2 \gamma \)
Substituting values of \( \alpha \), \( \beta \) and \( \gamma \), we get,
\( (n_2 - n_1) \frac{\text{arc PA}}{R} = n_1 (\frac{\text{arc PA}}{-u}) + n_2 (\frac{\text{arc PA}}{v}) \)
\( \therefore \frac{(n_2 - n_1)}{R} = \frac{n_2}{v} - \frac{n_1}{u} \)

Assumptions: To derive an expression that relates object and image distances with the radius of curvature for a point object, the two rays considered are assumed to be paraxial thus making the angles subtended by incident ray, normal and refracted ray with the principal axis very small.
In simple words: The expression \( \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \) relates object distance (u), image distance (v), refractive indices of two media (\( n_1, n_2 \)), and radius of curvature (R) for a spherical refracting surface. It is derived assuming paraxial rays, meaning all angles of incidence and refraction are small.

🎯 Exam Tip: Remember the sign conventions (Cartesian) and the paraxial approximation for small angles when deriving the formula for refraction at a spherical surface. This derivation is fundamental for understanding lens behavior.

Question 12. A spherical surface separates two transparent media. Derive an expression that relates object and image distances with the radius of curvature for a point object. Clearly state the assumptions, if any.
Answer:
i. Consider a spherical surface YPY' of radius curvature R, separating two transparent media of refractive indices n₁ and n₂ respectively with n₁ < n2.
ii. P is the pole and X'PX is the principal axis. A point object O is at a distance u from the pole, in the medium of refractive index n₁.
iii. In order to minimize spherical aberration, we consider two paraxial rays.
iv. The ray OP along the principal axis travels undeviated along PX. Another ray OA strikes the surface at A.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक गोलीय सतह को दिखाता है जो दो पारदर्शी माध्यमों को अलग करती है, जिसमें एक बिंदु वस्तु O से प्रकाश की किरणें सतह से अपवर्तित होने के बाद बिंदु I पर एक वास्तविक प्रतिबिंब बनाती हैं। मुख्य अक्ष, ध्रुव P, वक्रता केंद्र C और आपतन कोण (α, β, γ) प्रदर्शित हैं।
v. As n₁ < n2, the ray deviates towards the normal (CAN), travels along AZ and real image of point object O is formed at I.
vi. Let α, β and γ be the angles subtended by incident ray, normal and refracted ray with the principal axis.
∴ i = (α + β) and r = (β – γ)
vii. As, the rays are paraxial, all the angles can be considered to be very small.
i.e., sin i ≈ i and sin r ≈ r
Angles a, β and γ can also be expressed as,
\( \alpha = \frac{\text{arc PA}}{\text{OP}} = -\frac{\text{arc PA}}{u} \)
\( \beta = \frac{\text{arc PA}}{\text{PC}} = \frac{\text{arc PA}}{R} \)
and \( \gamma = \frac{\text{arc PA}}{\text{PI}} = \frac{\text{arc PA}}{v} \)
viii. According to Snell's law,
n₁ sin (i) = n2 sin (r)
For small angles, Snell's law can be written
as, n₁i = n2r
∴ n₁ (α + β) = n2 (β – γ)
∴ (n2 – n₁)β = n₁α + n2γ
Substituting values of α, β and γ, we get,
\( (n_2 - n_1) \frac{\text{arcPA}}{R} = n_1 \left( \frac{\text{arcPA}}{-u} \right) + n_2 \left( \frac{\text{arcPA}}{v} \right) \)

\( \implies \frac{(n_2-n_1)}{R} = \frac{n_2}{v} - \frac{n_1}{u} \)
Assumptions: To derive an expression that relates object and image distances with the radius of curvature for a point object, the two rays considered are assumed to be paraxial thus making the angles subtended by incident ray, normal and refracted ray with the principal axis very small.
In simple words: This derivation explains how light bends when passing through a spherical surface separating two different materials, showing the relationship between object distance, image distance, and the curvature of the surface, assuming rays are close to the axis.

🎯 Exam Tip: Understanding the Cartesian sign convention and the paraxial ray assumption is crucial for solving problems related to spherical surfaces. Pay attention to the refractive indices of the media.

Question 13. Derive lens makers' equation. Why is it called so? Under which conditions focal length f and radii of curvature R are numerically equal for a lens?
Answer:
i. Consider a lens of radii of curvature R1 and R2 kept in a medium such that refractive index of material of the lens with respect to the medium is denoted as n.
ii. Assuming the lens to be thin, P is the common pole for both the surfaces. O is a point object on the principal axis at a distance u from P.
iii. The refracting surface facing the object is considered as first refracting surface with radii R1.
iv. In the absence of second refracting surface, the paraxial ray OA deviates towards normal and would intersect axis at I₁. PI₁ = V₁ is the image distance for intermediate image I₁.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक पतले लेंस को दर्शाता है जिसकी अपवर्तनांक n है और यह एक माध्यम में रखा गया है। एक बिंदु वस्तु O से प्रकाश की किरणें पहली अपवर्तक सतह से अपवर्तित होकर एक मध्यवर्ती प्रतिबिंब I₁ बनाती हैं, जिससे लेंस मेकर सूत्र की व्युत्पत्ति को समझा जा सके।
v. For a curved surface,
\( \frac{(n_2-n_1)}{R} = \frac{n_2}{V} - \frac{n_1}{u} \)
Thus, in this case,
n2 = n, n₁ = 1, R = R₁, u = u and v = V1
∴ \( \frac{(n-1)}{R_1} = \frac{n}{V_1} - \frac{1}{u} \)....(1)
Before reaching I₁, the incident rays (AB and OP) strike the second refracting surface. In this case, image I₁ acts as a virtual object for second surface.
vii. For second refracting surface,
n2 = 1, n₁ = n, R = R2, u = v₁ and PI = v
∴ \( \frac{(1-n)}{R_2} = \frac{1}{V} - \frac{n}{V_1} \)
\( \implies -\frac{(n-1)}{R_2} = \frac{1}{V} - \frac{n}{V_1} \).....(2)
viii. Adding equations (1) and (2),
\( (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = \frac{1}{V} - \frac{1}{u} \)
For object at infinity, image is formed at focus, i.e., for u = ∞, v = f. Substituting this in above equation,
\( \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \).....(3)
This equation in known as the lens makers' formula.
ix. Since the equation can be used to calculate the radii of curvature for the lens, it is called the lens makers' equation.
x. The numeric value of focal length f and radius of curvature R is same under following two conditions:
Case I:
For a thin, symmetric and double convex lens made of glass (n = 1.5), R₁ is positive and R2 is negative but, |R1| = |R2].
In this case,
\( \frac{1}{f} = (1.5 - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) = 0.5 \left( \frac{2}{R} \right) = \frac{1}{R} \)
∴ f = R
Case II:
Similarly, for a thin, symmetric and double concave lens made of glass (n = 1.5), R₁ is negative and R2 is positive but, |R₁| = |R2].
In this case,
\( \frac{1}{f} = (1.5 - 1) \left( \frac{1}{-R} - \frac{1}{R} \right) = 0.5 \left( -\frac{2}{R} \right) = -\frac{1}{R} \)
∴ f = -R or |f| = |R|
In simple words: The lens maker's equation relates the focal length of a lens to its refractive index and the radii of curvature of its surfaces. It's used by lens manufacturers to design lenses with specific focal lengths, and in specific cases, the focal length can be numerically equal to the radius of curvature.

🎯 Exam Tip: Remember the sign conventions for radii of curvature (R1 and R2) for convex and concave surfaces when applying the lens maker's formula. Also, clearly state the assumptions of a thin lens and paraxial rays.

3. Answer The Following Questions In Detail.

Question 1. What are different types of dispersions of light? Why do they occur?
Answer:
i. There are two types of dispersions:
a. Angular dispersion
b. Lateral dispersion
ii. The refractive index of material depends on the frequency of incident light. Hence, for different colours, refractive index of material is different.
iii. For an obliquely incident ray, the angles of refraction are different for each colour and they separate as they travel along different directions resulting into angular dispersion.
iv. When a polychromatic beam of light is obliquely incident upon a plane parallel transparent slab, emergent beam consists of all component colours separated out.
v. In this case, these colours are parallel to each other and are also parallel to their initial direction resulting into lateral dispersion
In simple words: Dispersion of light is when white light splits into its constituent colors. This happens because the refractive index of a material varies with the color (frequency) of light, causing different colors to bend at different angles. Angular dispersion separates colors into different directions, while lateral dispersion just separates them spatially but keeps them parallel.

🎯 Exam Tip: Distinguish clearly between angular and lateral dispersion. Focus on the core reason: refractive index's dependence on wavelength/frequency for different colors.

Question 2. Define angular dispersion for a prism. Obtain its expression for a thin prism. Relate it with the refractive indices of the material of the prism for corresponding colours.
Answer:
i. If a polychromatic beam is incident upon a prism, the emergent beam consists of all the individual colours angularly separated. This phenomenon is known as angular dispersion for a prism.
ii. For any two component colours, angular dispersion is given by,
δ21 = δ2 – δ1
iii. For white light, we consider two extreme colours viz., red and violet.
∴ δVR = δV – δR
iv. For thin prism,
δ = A(n - 1)
δ21 = δ2 – δ1
= A(n2 – 1) – A(n₁ – 1) = A(n2 – n₁)
where n₁ and n2 are refractive indices for the two colours.
v. For white light,
δVR = δV – δR
= A(nV – 1) – A(nR – 1) = A(nV – nR).
In simple words: Angular dispersion in a prism is the difference in deviation angles for two different colors of light. For a thin prism, it's directly proportional to the prism's angle and the difference in refractive indices for those two colors, meaning different colors spread out at varying angles.

🎯 Exam Tip: Remember the formula for deviation in a thin prism, δ = A(n-1), as it's the foundation for deriving angular dispersion. Clearly state the relationship between angular dispersion and refractive indices.

Question 3. Explain and define dispersive power of a transparent material. Obtain its expressions in terms of angles of deviation and refractive indices.
Answer:
Ability of an optical material to disperse constituent colours is its dispersive power.
It is measured for any two colours as the ratio of angular dispersion to the mean deviation for those two colours. Thus, for the extreme colours of white light,
dispersive power is given by,
\( \omega = \frac{\delta_V - \delta_R}{(\frac{\delta_V + \delta_R}{2})} \approx \frac{\delta_V - \delta_R}{\delta_Y} = \frac{A(n_V - n_R)}{A(n_Y - 1)} = \frac{n_V - n_R}{n_Y - 1} \)
In simple words: Dispersive power is a material's ability to spread out different colors of light. It's defined as the ratio of the angular dispersion (how much the extreme colors separate) to the mean deviation (average bending) of light, and can also be expressed using the refractive indices for different colors.

🎯 Exam Tip: Clearly define dispersive power and memorize its two main forms: in terms of deviation angles and in terms of refractive indices. Pay attention to the subscripts (V for violet, R for red, Y for yellow/mean).

Question 4. (i) State the conditions under which a rainbow can be seen.
Answer:
A rainbow can be observed when there is a light shower with relatively large raindrop occurring during morning or evening time with enough sunlight around.
(ii) Explain the formation of a primary rainbow. For which angular range with the horizontal is it visible?
Answer:
i. A ray AB incident from Sun (white light) strikes the upper portion of a water drop at an incident angle i.
ii. On entering into water, it deviates and disperses into constituent colours. The figure shows the extreme colours (violet and red).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक प्राथमिक इंद्रधनुष के निर्माण को दर्शाता है। इसमें सूर्य का प्रकाश एक पानी की बूंद पर आपतित होता है, अपवर्तित और विक्षेपित होता है, फिर आंतरिक रूप से परावर्तित होता है, और अंत में प्रेक्षक की ओर बाहर निकलता है, जिससे लाल और बैंगनी रंगों के लिए न्यूनतम विचलन कोण दिखते हैं।
iii. Refracted rays BV and BR strike the opposite inner surface of water drop and suffer internal reflection.
iv. These reflected rays finally emerge from V' and R' and can be seen by an observer on the ground.
v. For the observer they appear to be coming from opposite side of the Sun.
vi. Minimum deviation rays of red and violet colour are inclined to the ground level at θR = 42.8° ≈ 43° and θV = 40.8 ≈ 41° respectively. As a result, in the rainbow, the red is above and violet is below.
(iii) Explain the formation of a secondary rainbow. For which angular range with the horizontal is it visible?
Answer:
i. A ray AB incident from Sun (white light) strikes the lower portion of a water drop at an incident angle i.
ii. On entering into water, it deviates and disperses into constituent colours. The figure shows the extreme colours (violet and red).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक द्वितीयक इंद्रधनुष के निर्माण को दर्शाता है। इसमें सूर्य का प्रकाश एक पानी की बूंद पर आपतित होता है, अपवर्तन और विक्षेपण के बाद दो बार आंतरिक परावर्तन से गुजरता है, और फिर प्रेक्षक की ओर बाहर निकलता है, जिससे बैंगनी और लाल रंगों के लिए न्यूनतम विचलन कोण दिखते हैं।
iii. Refracted rays BV and BR finally emerge the drop from V' and R' after suffering two internal reflections and can be seen by an observer on the ground.
iv. Minimum deviation rays of red and violet colour are inclined to the ground level at θR ≈ 51° and θV ≈ 53° respectively. As a result, in the rainbow, the violet is above and red is below.
(iv) Is it possible to see primary and secondary rainbow simultaneously? Under what conditions?
Answer:
Yes, it is possible to see primary and secondary rainbows simultaneously. This can occur when the centres of both the rainbows coincide.
In simple words: Rainbows form when sunlight interacts with raindrops, involving refraction, dispersion, and total internal reflection. A primary rainbow involves one internal reflection with red on top, while a secondary rainbow involves two internal reflections with violet on top. Both can be seen together if their centers align, typically during morning or evening light showers.

🎯 Exam Tip: Differentiate clearly between primary and secondary rainbows based on the number of internal reflections (one vs. two) and the order of colors. Remember the specific angular ranges and the weather conditions required.

Question 5. (i) Explain chromatic aberration for spherical lenses. State a method to minimize or eliminate it.
Answer:
Lenses are prepared by using a transparent material medium having different refractive index for different colours. Hence angular dispersion is present.
If the lens is thick, this will result into notably different foci corresponding to each colour for a polychromatic beam, like a white light. This defect is called chromatic aberration.
As violet light has maximum deviation, it is focussed closest to the pole.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र उत्तल और अवतल लेंसों में वर्ण विपथन को दर्शाता है। सफेद प्रकाश लेंस से गुजरने पर अपने घटक रंगों में विक्षेपित होता है, जिससे प्रत्येक रंग के लिए अलग-अलग फोकल बिंदु (जैसे लाल के लिए FR और बैंगनी के लिए FV) बनते हैं, जिसके परिणामस्वरूप एक धुंधली छवि बनती है।
Reducing/eliminating chromatic aberration:
1. Eliminating chromatic aberrations for all colours is impossible. Hence, it is minimised by eliminating aberrations for extreme colours.
2. This is achieved by using either a convex and a concave lens in contact or two thin convex lenses with proper separation. Such a combination is called achromatic combination.
(ii) What is achromatism? Derive a condition to achieve achromatism for a lens combination. State the conditions for it to be converging.
Answer:
i. To eliminate chromatic aberrations for extreme colours from a lens, either a convex and a concave lens in contact or two thin convex lenses with proper separation are used.
ii. This combination is called achromatic combination. The process of using this combination is termed as achromatism of a lens.
iii. Let ω₁ and ω₂ be the dispersive powers of materials of the two component lenses used in contact for an achromatic combination.
iv. Let V, R and Y denote the focal lengths for violet, red and yellow colours respectively.
v. For lens 1, let
\( K_1 = \left( \frac{1}{R_1} - \frac{1}{R_2} \right)_1 \) and \( K_2 = \left( \frac{1}{R_1} - \frac{1}{R_2} \right)_2 \)
vi. For the combination to be achromatic, the resultant focal length of the combination must be the same for both the colours,
∴ fV = fR
\( \implies \frac{1}{f_V} = \frac{1}{f_R} \)
For two thin lenses in contact, \( \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} \)
\( \implies \frac{1}{(f_V)_1} + \frac{1}{(f_V)_2} = \frac{1}{(f_R)_1} + \frac{1}{(f_R)_2} \)
vii. From Lens maker's formula,
[(nV)₁ -1]K₁ +[(nV)₂-1]K₂
= [(nR)₁ -1]K₁ +[(nR)₂-1]K₂
\( \implies \frac{K_1}{K_2} = - \frac{(n_V)_2 - (n_R)_2}{(n_V)_1 - (n_R)_1} \)....(1)
viii. Similarly, for mean colour (yellow),
\( \frac{1}{f_Y} = \frac{1}{(f_Y)_1} + \frac{1}{(f_Y)_2} \)....(2)
\( \implies \frac{K_1}{K_2} = - \frac{(n_Y)_2 - 1}{(n_Y)_1 - 1} \frac{(f_Y)_1}{(f_Y)_2} \)....(3)
ix. From equations (1) and (3),
\( \frac{(n_V)_2 - (n_R)_2}{(n_Y)_2 - 1} \times \frac{(n_Y)_1 - 1}{(n_V)_1 - (n_R)_1} = - \frac{(f_Y)_1}{(f_Y)_2} \)
Now, dispersive power \( \omega_1 = \frac{(n_V)_1 - (n_R)_1}{(n_Y)_1 - 1} \) and \( \omega_2 = \frac{(n_V)_2 - (n_R)_2}{(n_Y)_2 - 1} \)
x. Substituting values of ω₁ and ω₂ in equation (4), we get,
\( \frac{\omega_2}{\omega_1} = - \frac{(f_Y)_1}{(f_Y)_2} \)
This is the condition for achromatism of a combination of lenses.
Condition for converging:
For this combination to be converging, fY must be positive.
Using equation (3), for fY to be positive, (fY)1 < (fY)2 \( \implies \) ω₁ < ω2
In simple words: Chromatic aberration is a lens defect where different colors of light focus at different points, leading to fuzzy images. It can be minimized by combining a convex and a concave lens (called an achromatic combination) to ensure all colors focus at roughly the same point. Achromatism is the process of achieving this color correction.

🎯 Exam Tip: For chromatic aberration, remember it's due to the dependence of refractive index on color. For achromatism, focus on the condition \( \frac{\omega_1}{f_1} + \frac{\omega_2}{f_2} = 0 \) or similar forms, and the implication for a converging system (ratio of focal lengths and dispersive powers).

Question 6. Describe spherical aberration for spherical lenses. What are different ways to minimize or eliminate it?
Answer:
i. All the formulae used for image formation by lenses are based on some assumption. However, in reality these assumptions are not always true.
ii. A single point focus in case of lenses is possible only for small aperture spherical lenses and for paraxial rays.
iii. The rays coming from a distant object farther from principal axis no longer remain parallel to the axis. Thus, the focus gradually shifts towards pole.
iv. This defect arises due to spherical shape of the refracting surface, hence known as spherical aberration. It results into a blurred image with unclear boundaries.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र उत्तल और अवतल लेंसों में गोलीय विपथन को दर्शाता है। इसमें दिखाया गया है कि कैसे लेंस के किनारे से आने वाली किरणें (FM) और मुख्य अक्ष के करीब से आने वाली किरणें (FP) अलग-अलग बिंदुओं पर केंद्रित होती हैं, जिससे एक स्पष्ट बिंदु फोकस के बजाय एक धुंधली छवि बनती है।
v. As shown in figure, the rays near the edge of the lens converge at focal point FM. Whereas, the rays near the principal axis converge at point FP. The distance between FM and FP is measured as the longitudinal spherical aberration.
vi. In absence of this aberration, a single point image can be obtained on a screen. In the presence of spherical aberration, the image is always a circle.
vii. At a particular location of the screen (across AB in figure), the diameter of this circle is minimum. This is called the circle of least confusion. Radius of this circle is transverse spherical aberration.
Methods to eliminate/reduce spherical aberration in lenses:
i. Cheapest method to reduce the spherical aberration is to use a planoconvex or planoconcave lens with curved side facing the incident rays.
ii. Certain ratio of radii of curvature for a given refractive index almost eliminates the spherical aberration. For n = 1.5, the ratio is
\( \frac{R_1}{R_2} = \frac{1}{6} \) and for n = 2, \( \frac{R_1}{R_2} = \frac{1}{5} \)
iii. Use of two thin converging lenses separated by distance equal to difference between their focal lengths with lens of larger focal length facing the incident rays considerably reduces spherical aberration.
iv. Spherical aberration of a convex lens is positive (for real image), while that of a concave lens is negative. Thus, a suitable combination of them can completely eliminate spherical aberration.
In simple words: Spherical aberration is a lens defect where light rays passing through different parts of a spherical lens (center vs. edges) do not converge at a single focal point, causing a blurred image. It's caused by the lens's spherical shape and can be reduced by using plano-convex/concave lenses, specific lens curvature ratios, or combining multiple lenses.

🎯 Exam Tip: Focus on explaining *why* spherical aberration occurs (spherical shape leading to different focal points for paraxial and marginal rays) and then listing specific, actionable remedies like using plano-lenses or combinations of lenses.

Question 7. Define and describe magnifying power of an optical instrument. How does it differ from linear or lateral magnification?
Answer:
i. Angular magnification or magnifying power of an optical instrument is defined as the ratio of the visual angle made by the image formed by that optical instrument (β) to the visual angle subtended by the object when kept at the least distance of distinct vision (α).
ii. The linear magnification is the ratio of the size of the image to the size of the object.
iii. When the distances of the object and image formed are very large as compared to the focal lengths of the instruments used, the magnification becomes infinite. Whereas, the magnifying power being the ratio of angle subtended by the object and image, gives the finite value.
iv. For example, in case of a compound microscope,
Mmin = \( \frac{D}{f} \) = \( \frac{25}{5} \) = 5 and Mmax = \( 1 + \frac{D}{f} \) = 6
Hence image appears to be only 5 to 6 times bigger for a lens of focal length 5 cm.
For Mmin = \( \frac{D}{f} \) = 5, V = ∞
∴ Lateral magnification (m) = \( \frac{v}{u} \) = ∞
Thus, the image size is infinite times that of the object, but appears only 5 times bigger.
In simple words: Magnifying power, or angular magnification, is about how much larger an object appears to the eye through an optical instrument, based on the visual angle. Linear magnification, on the other hand, is a simple ratio of the image's physical size to the object's physical size. Angular magnification is more useful for distant objects as it provides a finite value even when linear magnification approaches infinity.

🎯 Exam Tip: Clearly define both angular and linear magnification and highlight their key difference: angular magnification considers the visual angle, making it useful for distant objects and preventing infinite values, unlike linear magnification in certain cases.

Question 8. Derive an expression for magnifying power of a simple microscope. Obtain its minimum and maximum values in terms of its focal length.
Answer:
i. Figure (a) shows visual angle α made by an object, when kept at the least distance of distinct vision (D = 25 cm). Without an optical instrument this is the greatest possible visual angle as we cannot get the object closer than this.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र (a) न्यूनतम स्पष्ट दृष्टि दूरी D पर रखी वस्तु AB द्वारा आँख पर बनाया गया दृश्य कोण α दिखाता है, जो एक ऑप्टिकल उपकरण के बिना वस्तु का सबसे बड़ा दृश्य कोण है।
ii. Figure (b) shows a convex lens forming erect, virtual and magnified image of the same object, when placed within the focus.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र (b) एक उत्तल लेंस द्वारा वस्तु AB का सीधा, आभासी और आवर्धित प्रतिबिंब दिखाता है, जब वस्तु को लेंस के फोकस (F) के भीतर रखा जाता है।
iii. The visual angle β of the object and the image in this case are the same.
However, this time the viewer is looking at the image which is not closer than D. Hence the same object is now at a distance smaller than D.
iv. Angular magnification or magnifying power, in this case, is given by
M = \( \frac{\text{Visual angle of the image}}{\text{Visual angle of the object at D}} = \frac{\beta}{\alpha} \)
For small angles,
M = \( \frac{\beta}{\alpha} = \frac{\tan(\beta)}{\tan(\alpha)} = \frac{\text{BA/PA}}{\text{BA/D}} = \frac{D}{u} \)
v. For maximum magnifying power, the image should be at D. For thin lens, considering thin lens formula.
\( \frac{1}{f} = \frac{1}{V} - \frac{1}{u} \)
In case of simple microscope,
f = +f, v = - D, u = -u and M = Mmax
\( \frac{1}{f} = \frac{1}{-D} - \frac{1}{-u} \)
\( \implies \frac{D}{f} = \frac{D}{-D} + \frac{D}{u} \)
\( \implies \frac{D}{f} = -1 + \frac{D}{u} \)
As, M = \( \frac{D}{u} \)
Mmax = \( 1 + \frac{D}{f} \)
vi. For minimum magnifying power, v = ∞ and
u = f (numerically)
∴ Mmin = \( \frac{D}{u} = \frac{D}{f} \)
In simple words: A simple microscope uses a convex lens to produce a magnified image. Its magnifying power is the ratio of the visual angle with the lens to the visual angle without it. Maximum magnification occurs when the image is at the least distance of distinct vision, while minimum magnification occurs when the image is at infinity, defined by the lens's focal length.

🎯 Exam Tip: Clearly state the two conditions for maximum and minimum magnifying power (image at D and image at infinity). Remember the formulas \( M_{\text{max}} = 1 + \frac{D}{f} \) and \( M_{\text{min}} = \frac{D}{f} \), and the definition of D (least distance of distinct vision).

Question 9. Derive the expressions for the magnifying power and the length of a compound microscope using two convex lenses.
Answer:
i. The final image formed in compound microscope (A” B") as shown in figure, makes a visual angle β at the eye.
Visual angle made by the object from distance D is α.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक संयुक्त सूक्ष्मदर्शी की किरण आरेख को दर्शाता है, जिसमें दो उत्तल लेंस - अभिदृश्यक लेंस (objective lens) और नेत्रिका लेंस (eye lens) - का उपयोग करके एक वस्तु AB का आवर्धित अंतिम प्रतिबिंब A"B" बनता है। इसमें विभिन्न दूरियाँ और कोणों को दर्शाया गया है।
From figure,
\( \tan \beta = \frac{A''B''}{V_e} = \frac{A'B'}{u_e} \)
and \( \tan \alpha = \frac{AB}{D} \)
In simple words: A compound microscope uses two convex lenses, an objective and an eyepiece, to achieve high magnification. The magnifying power is the product of the magnifications of both lenses, and its length is the sum of the image distance from the objective and the object distance for the eyepiece.

🎯 Exam Tip: Remember that the total magnifying power of a compound microscope is the product of the magnifications of the objective and eyepiece. For the length, it's approximately the sum of focal lengths for normal adjustment.

Question 10. What is a terrestrial telescope and an astronomical telescope?
Answer:
1. Telescopes used to see the objects on the Earth, like mountains, trees, players playing a match in a stadium, etc. are called terrestrial telescopes.
2. In such case, the final image must be erect. Eye lens used for this purpose must be concave and such a telescope is popularly called a binocular.
3. Most of the binoculars use three convex lenses with proper separation. The image formed by second lens is inverted with respect to object. The third lens again inverts this image and makes final image erect with respect to the object.
4. An astronomical telescope is the telescope used to see the objects like planets, stars, galaxies, etc. In this case there is no necessity of erect image. Such telescopes use convex lens as eye lens.
In simple words: A terrestrial telescope is for viewing objects on Earth, requiring an erect final image, often achieved with extra lenses or a prism system (like binoculars). An astronomical telescope is for observing celestial bodies and does not require an erect image, typically using a convex eyepiece.

🎯 Exam Tip: The main distinction between terrestrial and astronomical telescopes lies in the requirement for an erect (terrestrial) vs. inverted (astronomical) final image, which dictates the lens configuration.

Question 11. Obtain the expressions for magnifying power and the length of an astronomical telescope under normal adjustments.
Answer:
i. For telescopes, α is the visual angle of the object from its own position, which is practically at infinity.
ii. Visual angle of the final image is β and its position can be adjusted to be at D. However, under normal adjustments, the final image is also at infinity making a greater visual angle than that of the object.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र सामान्य समायोजन में एक खगोलीय दूरबीन को दर्शाता है, जिसमें एक अभिदृश्यक लेंस और एक नेत्रिका लेंस का उपयोग करके एक दूर की वस्तु का आवर्धित और अंतिम प्रतिबिंब अनंत पर बनता है, और विभिन्न फोकल लंबाई (fo, fe) तथा दृश्य कोण (α, β) दिखाए गए हैं।
iii. The parallax at the cross wires can be avoided by using the telescopes in normal adjustments.
iv. Objective of focal length fo focusses the parallel incident beam at a distance fo from the objective giving an inverted image AB.
v. For normal adjustment, the intermediate image AB forms at the focus of the eye lens. Rays refracted beyond the eye lens form a parallel beam inclined at an angle β with the principal axis.
vi. Angular magnification or magnifying power for telescope is given by,
M = \( \frac{\beta}{\alpha} = \frac{\tan(\beta)}{\tan(\alpha)} = \frac{\text{BA/PB}}{\text{BA/POB}} = \frac{f_o}{f_e} \)
vii. Length of the telescope for normal adjustment is, L = fo + fe.
In simple words: An astronomical telescope uses two convex lenses to view distant objects, creating an inverted image. Under normal adjustment, both the intermediate and final images are at infinity. Its magnifying power is the ratio of the objective lens's focal length to the eyepiece's focal length, and its length is the sum of these focal lengths.

🎯 Exam Tip: For astronomical telescopes, remember that normal adjustment means the final image is at infinity, simplifying the length and magnifying power calculations. The key is \( M = \frac{f_o}{f_e} \) and \( L = f_o + f_e \).

Question 12. What are the limitations in increasing the magnifying powers of (i) simple microscope (ii) compound microscope (iii) astronomical telescope?
Answer:
(i) In case of simple microscope
\( M_{max} = \frac{D}{u} = 1 + \frac{D}{f} \)
Thus, the limitation in increasing the magnifying power is determined by the value of focal length and the closeness with which the lens can be held near the eye.
(ii) In case of compound microscope,
\( M = m_o \times M_e = \frac{v_o}{u_o} \times \frac{D}{u_e} \)
Thus, in order to increase mo, we need to decrease uo. Thereby, the object comes closer and closer to the focus of the objective. This increases vo and hence length of the microscope. Therefore, mQ can be increased only within the limitation of length of the microscope.
(iii) In case of telescopes,
\( M = \frac{f_o}{f_e} \)
Where \( f_o \) = focal length of the objective
\( f_e \) = focal length of the eye-piece
Length of the telescope for normal adjustment is, \( L = f_o + f_e \).
Thus, magnifying power of telescope can be increased only within the limitations of length of the telescope.
In simple words: The magnifying power of optical instruments like microscopes and telescopes is limited by their focal lengths, object distances, and physical lengths, preventing unlimited magnification.

🎯 Exam Tip: Understanding the formulas for magnifying power and the physical constraints for each optical instrument is crucial for scoring well in such derivation and explanation questions.

4. Solve The Following Numerical Examples

Question 1. A monochromatic ray of light strikes the water (n = 4/3) surface in a cylindrical vessel at angle of incidence 53°. Depth of water is 36 cm. After striking the water surface, how long will the light take to reach the bottom of the vessel? [Angles of the most popular Pythagorean triangle of sides in the ratio 3 : 4 : 5 are nearly 37°, 53° and 90°]
Answer:
From figure, ray PO = incident ray
ray OA = refracted ray
QOB = Normal to the water surface.
Given that,
\( \angle POQ \) = angle of incidence (\( \theta_1 \)) = 53°
Seg OB = 36 cm and \( n_{water} = \frac{4}{3} \)
From Snell's law,
\( n_1 \sin \theta_1 = n_2 \sin \theta_2 \)

\( \implies n_{water} = \frac{\sin \theta_1}{\sin \theta_2} \)

\( \implies \sin \theta_2 = \frac{\sin \theta_1}{n_{water}} = \frac{\sin(53°) \times 3}{4} \)

\( \implies \theta_2 \sim 37° \)
\( \triangle OBA \) forms a Pythagorean triangle with angles 53°, 37° and 90°.
Thus, sides of \( \triangle OBA \) will be in ratio 3 : 4 : 5 Such that OA forms the hypotenuse.
From figure, we can infer that,
Seg OB = 4x = 36 cm
\( \implies \) x = 9
\( \implies \) seg OA = 5x = 45 cm and
seg AB = 3x = 27 cm.
This means the light has to travel 45 cm to reach the bottom of the vessel.
The speed of the light in water is given by,
\( v = \frac{c}{n} \)

\( \implies v = \frac{3 \times 10^8}{4/3} = \frac{9}{4} \times 10^8 \) m/s

\( \implies \) Time taken by light to reach the bottom of vessel is,
\( t = \frac{s}{v} = \frac{45 \times 10^{-2}}{9 \times 10^8} = 20 \times 10^{-10} = 2 \) ns or 0.002 µs
In simple words: Using Snell's law and trigonometric properties of a Pythagorean triangle, we find the path length of light in water and then calculate the time taken using the speed of light in water.

🎯 Exam Tip: Remember to correctly apply Snell's law and convert units. Identifying geometric properties like Pythagorean triangles can simplify path length calculations.

Question 2. Estimate the number of images produced if a tiny object is kept in between two plane mirrors inclined at 35°, 36°, 40° and 45°.
Answer:
(i) For \( \theta_1 = 35° \)
\( n_1 = \frac{360}{\theta_1} = \frac{360}{35} = 10.28 \)
As n1 is non-integer, \( N_1 \) = integral part of \( n_1 \) = 10
(ii) For \( \theta_2 = 36° \)
\( n_2 = \frac{360}{36} = 10 \)
As n2 is even integer, \( N_2 = (n_2 - 1) = 9 \)
(iii) For \( \theta_3 = 40° \)
\( n_3 = \frac{360}{40} = 9 \)
As n3 is odd integer.
Number of images seen (N3) = n3 - 1 = 8
(if the object is placed at the angle bisector) or Number of images seen (N3) = n3 = 9
(if the object is placed off the angle bisector)
(iv) For \( \theta_4 = 45° \)
\( n_4 = \frac{360}{45} = 8 \)
As n4 is even integer,
\( N_4 = n_4 - 1 = 7 \)
In simple words: The number of images formed by two inclined plane mirrors depends on the angle between them and whether \( 360/\theta \) is an even, odd, or non-integer value.

🎯 Exam Tip: Always remember the rules for calculating the number of images based on whether \( 360/\theta \) is even, odd, or non-integer, and consider the special case when the object is on the angle bisector for odd \( 360/\theta \).

Question 3. A rectangular sheet of length 30 cm and breadth 3 cm is kept on the principal axis of a concave mirror of focal length 30 cm. Draw the image formed by the mirror on the same diagram, as far as possible on scale.
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक अवतल दर्पण के सामने रखी 30 सेमी लंबाई और 3 सेमी चौड़ाई वाली आयताकार शीट द्वारा बनी छवि को दर्शाता है। शीट का एक सिरा (D) 90 सेमी पर है, जबकि दूसरा सिरा (A) 60 सेमी पर है। दर्पण का केंद्र वक्रता (C) 60 सेमी पर है और फोकस (F) 30 सेमी पर है। किरणें दर्पण से परावर्तित होकर शीट की उलटी और वास्तविक छवि (D'Q'A'B') बनाती हैं।
[Note: The question has been modified and the ray digram is inserted in question in order to find the correct position of the image.]
In simple words: For a concave mirror, an object placed beyond the center of curvature forms a real, inverted, and diminished image, which can be graphically depicted using ray tracing.

🎯 Exam Tip: For ray diagrams, ensure rays pass through or appear to pass through focal points and the center of curvature for accurate image formation. Labeling all points and distances clearly is vital.

Question 4. A car uses a convex mirror of curvature 1.2 m as its rear-view mirror. A minibus of cross section 2.2 m x 2.2 m is 6.6 m away from the mirror. Estimate the image size
Answer:
For a convex mirror,
\( f = + \frac{R}{2} = + \frac{1.2}{2} = +0.6 \) m
Given that, a minibus, approximately of a shape of square is at distance 6.6 m from mirror.
i.e., u = -6.6 m

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक उत्तल दर्पण और उसके सामने रखे एक मिनीबस को दर्शाता है। मिनीबस दर्पण से 6.6 मीटर की दूरी पर है। दर्पण का फोकस (F) 0.6 मीटर पर है, और यह चित्र मिनीबस द्वारा बनने वाली आभासी और छोटी छवि को दर्शाने के लिए किरण आरेख का उपयोग करता है।
From mirror formula,
\( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \)

\( \implies \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \)

\( \implies \frac{1}{v} = \frac{1}{0.6} - \frac{1}{(-6.6)} \)

\( \implies \frac{1}{v} = \frac{1}{0.6} + \frac{1}{6.6} = \frac{6.6 + 0.6}{0.6 \times 6.6} = \frac{7.2}{3.96} = \frac{20}{11} \)

\( \implies v = 0.55 \) m
As, lateral magnification,
\( m = \frac{h_2}{h_1} = - \frac{v}{u} \)

\( \implies \frac{h_2}{(2.2)} = - \frac{(0.55)}{(-6.6)} \)

\( \implies h_2 = \frac{0.55 \times 2.2}{6.6} \)

\( \implies h_2 = \frac{0.55}{3} \)

\( \implies h_2 = 0.183 \) m
i.e., \( h_2 \approx 0.2 \) m
In simple words: For a convex rear-view mirror, the image is always virtual, erect, and diminished. Using the mirror formula and magnification formula, we can calculate the image distance and height.

🎯 Exam Tip: Remember to use the correct sign conventions for focal length (positive for convex), object distance (negative for real objects), and image distance. Pay attention to unit consistency throughout the calculation.

Question 5. A glass slab of thickness 2.5 cm having refractive index 5/3 is kept on an ink spot. A transparent beaker of very thin bottom, containing water of refractive index 4/3 up to 8 cm, is kept on the glass block. Calculate apparent depth of the ink spot when seen from the outside air.
Answer:
When observed from the outside air, the light coming from ink spot gets refracted twice; once through glass and once through water.
\( \implies \) When observed from water,
\( \frac{n_g}{n_w} = \frac{\text{Real depth}}{\text{Apparent depth}} \)
\( \implies \frac{5/3}{4/3} = \frac{2.5}{\text{Apparent depth}} \)
\( \implies \frac{5}{4} = \frac{2.5}{\text{Apparent depth}} \)

\( \implies \text{Apparent depth} = \frac{2.5 \times 4}{5} = 2 \) cm
Now when observed from outside air, the total real depth of ink spot can be taken as (8 + 2) cm = 10 cm.
\( \frac{n_w}{n_{air}} = \frac{\text{Real depth}}{\text{Apparent depth}} \)

\( \implies \text{Apparent depth} = \frac{10}{4/3} \)

\( \implies \text{Apparent depth} = \frac{10 \times 3}{4} = 7.5 \) cm
In simple words: The apparent depth of an object beneath multiple layers of different refractive indices is found by calculating the apparent depth sequentially through each medium until reaching the observer's medium.

🎯 Exam Tip: When dealing with multiple layers, calculate apparent depth step-by-step from the object towards the observer, using the apparent position from the previous layer as the "real depth" for the next refraction. Ensure correct refractive indices are used for each interface.

Question 64. A convex lens held some distance above a 6 cm long pencil produces its image of SOME size. On shifting the lens by a distance equal to its focal length, it again produces the image of the SAME size as earlier. Determine the image size.
Answer:
For a convex lens, it is given that the image size remains unchanged after shifting the lens through distance equal to its focal length. From given conditions, it can be inferred that the object distance should be \( u = - \frac{f}{2} \)
Also, \( h_1 = 6 \) cm
From formula for thin lenses,
\( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)

\( \implies \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \)

\( \implies \frac{1}{v} = \frac{1}{f} + \frac{1}{(-f/2)} = \frac{1}{f} - \frac{2}{f} = - \frac{1}{f} \)

\( \implies v = -f \)
Now, magnification of the lens is,
\( m = \frac{h_2}{h_1} = - \frac{v}{u} = - \frac{(-f)}{(-f/2)} = -2 \)

\( \implies h_2 = h_1 \times |-2| = 6 \times 2 = 12 \) cm
In simple words: For a convex lens, if the object is placed at half the focal length (f/2) from the lens, it produces a virtual image twice the size. When the lens is shifted by its focal length, maintaining the same object size means the object must be at f/2.

🎯 Exam Tip: For lens problems, correctly apply the lens formula and magnification formula, paying close attention to sign conventions for object/image distances and focal length. The condition of unchanged image size after shifting the lens implies specific object placement relative to the focal length.

Question 7. Figure below shows the section ABCD of a transparent slab. There is a tiny green LED light source at the bottom left corner B. A certain ray of light from B suffers total internal reflection at nearest point P on the surface AD and strikes the surface CD at point Q. Determine refractive index of the material of the slab and distance DQ. At Q, the ray PQ will suffer partial or total internal reflection?
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक पारदर्शी स्लैब ABCD को दर्शाता है जिसमें निचले बाएं कोने B पर एक LED प्रकाश स्रोत है। B से एक प्रकाश किरण सतह AD पर बिंदु P पर पूर्ण आंतरिक परावर्तन से गुजरती है और सतह CD पर बिंदु Q से टकराती है। कोणों और दूरियों को 3 सेमी, 4 सेमी और 6 सेमी के रूप में चिह्नित किया गया है, जो प्रकाश के मार्ग और परावर्तन को समझने में मदद करता है।
As, the light ray undergo total internal reflection at P, the ray BP may be incident at critical angle.
For a Pythagorean triangle with sides in ratio 3 : 4 : 5 the angle opposite to side 3 units is 37° and that opposite to 4 units is 53°.
Thus, from figure, we can say, in \( \triangle BAP \)
\( \angle ABP = 53° \)
\( \angle BPN = i_c = 53° \)

\( \implies n_{glass} = \frac{1}{\sin i_c} = \frac{1}{\sin(53°)} \approx \frac{1}{0.8} = \frac{5}{4} \)

\( \implies \) Refractive index (n) of the slab is \( \frac{5}{4} \)
From symmetry, \( \triangle PDQ \) is also a Pythagorean triangle with sides in ratio QD : PD : PQ = 3 : 4 : 5.
PD = 2 cm \( \implies \) QD = 1.5 cm.
As critical angle is \( i_c = 53° \) and angle of incidence at Q, \( \angle PQN = 37° \) is less than critical angle, there will be partial internal reflection at Question
In simple words: By applying the conditions for total internal reflection and using trigonometric properties of the given geometry, we can determine the refractive index and predict whether light will undergo partial or total internal reflection.

🎯 Exam Tip: Clearly identify the critical angle using the relationship \( n = 1/\sin(i_c) \). Use geometric properties and angle calculations accurately to determine incidence angles at different surfaces to predict reflection/refraction outcomes.

Question 8. A point object is kept 10 cm away from a double convex lens of refractive index 1.5 and radii of curvature 10 cm and 8 cm. Determine location of the final image considering paraxial rays only.
Answer:
Given that, \( R_1 = 10 \) cm, \( R_2 = -8 \) cm,
\( u = -10 \) cm and \( n = 1.5 \)
From lens maker's equation,
\( \frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \)

\( \implies \frac{1}{f} = (1.5-1) \left( \frac{1}{10} - \frac{1}{(-8)} \right) \)

\( \implies \frac{1}{f} = 0.5 \times \left( \frac{1}{10} + \frac{1}{8} \right) \)

\( \implies \frac{1}{f} = 0.5 \times \frac{4+5}{40} = 0.5 \times \frac{9}{40} = \frac{4.5}{40} = \frac{9}{80} \)

\( \implies f = \frac{80}{9} \) cm
Now,
\( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)

\( \implies \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \)

\( \implies \frac{1}{v} = \frac{9}{80} + \frac{1}{(-10)} = \frac{9}{80} - \frac{8}{80} = \frac{1}{80} \)

\( \implies v = 80 \) cm
In simple words: The focal length of a lens can be determined using the lens maker's formula from its refractive index and radii of curvature. Then, the image position is found using the thin lens formula.

🎯 Exam Tip: Pay close attention to sign conventions for \( R_1 \) and \( R_2 \) for convex/concave surfaces in the lens maker's formula. A common mistake is to forget the negative sign for \( u \) (real object) and for \( R_2 \) (second surface for a double convex lens where light emerges from the convex side, meaning its center of curvature is on the other side). Double-check arithmetic carefully.

Question 9. A monochromatic ray of light is incident at 37° on an equilateral prism of refractive index 3/2. Determine angle of emergence and angle of deviation. If angle of prism is adjustable, what should its value be for emergent ray to be just possible for the same angle of incidence.
Answer:
By Snell's law, in case of prism,
\( n = \frac{\sin (i)}{\sin (r)} \)

\( \implies \frac{3}{2} = \frac{\sin (37°)}{\sin (r_1)} \)

\( \implies r_1 = \sin^{-1} \left( \frac{0.6018}{3/2} \right) \)

\( \implies r_1 = \sin^{-1} (0.4012) \)

\( \implies r_1 = 23°39' \)
For equilateral prism, A = 60°
Also, \( A = r_1 + r_2 \)
\( \implies r_2 = A - r_1 = 60° - 23°39' = 36°21' \)
Applying Snell's law on the second surface of prism,
\( \frac{1}{n} = \frac{\sin (r_2)}{\sin (e)} \)

\( \implies \frac{2}{3} = \frac{\sin (36°21')}{\sin (e)} \)

\( \implies \sin (e) = \frac{0.5927}{2/3} \)

\( \implies e = \sin^{-1} (0.889) \)

\( \implies e = 62°44' \)

\( \implies e \approx 63° \)
For any prism,
\( i + e = A + \delta \)

\( \implies \delta = (i + e) - A \)

\( \implies \delta = (37 + 63) - 60 \)

\( \implies \delta = 40° \)
For an emergent ray to just emerge, the angle \( r'_2 \) acts as a critical angle.
\( \implies r'_2 = \sin^{-1} \left( \frac{1}{n} \right) = \sin^{-1} \left( \frac{2}{3} \right) \)

\( \implies r'_2 = 41°48' \)
As, \( A' = r'_1 + r'_2 \) and i to be kept the same.
\( \implies A' = 23°39' + 41°48' \)

\( \implies A' = 65°27' \)
In simple words: The path of light through a prism involves two refractions, allowing us to calculate the angles of incidence, refraction, emergence, and deviation using Snell's law and the prism angle. For the emergent ray to be just possible, the angle of incidence at the second face must be the critical angle.

🎯 Exam Tip: Remember the prism formulas for Snell's law at both surfaces (\( i, r_1 \) and \( r_2, e \)) and the relationship \( A = r_1 + r_2 \). For the critical emergence condition, set the second angle of incidence (\( r_2 \)) equal to the critical angle (\( i_c \)).

Question 10. From the given data set, determine angular dispersion by the prism and dispersive power of its material for extreme colours. \( n_R = 1.62 \) \( n_V = 1.66 \), \( \delta_R = 3.1° \)
Answer:
Given: \( n_R = 1.62 \), \( n_V = 1.66 \), \( \delta_R = 3.1° \)
To find:
(i) Angular dispersion (\( \delta_{VR} \))
(ii) Dispersive power (\( \omega_{VR} \))
Formula:
(i) \( \delta = A (n - 1) \)
(ii) \( \delta_{VR} = \delta_V - \delta_R \)
(iii) \( \omega = \frac{\delta_V - \delta_R}{(\delta_V + \delta_R)/2} \)
Calculation: From formula (i),
\( \delta_R = A(n_R - 1) \)

\( \implies A = \frac{\delta_R}{(n_R - 1)} = \frac{3.1}{(1.62 - 1)} = \frac{3.1}{0.62} = 5 \)
\( \delta_V = A(n_V - 1) = 5 \times (1.66 - 1) = 5 \times 0.66 = 3.3° \)
From formula (ii),
\( \delta_{VR} = 3.3 - 3.1 = 0.2° \)
From formula (iii),
\( \omega_{VR} = \frac{3.3 - 3.1}{(3.3 + 3.1)/2} = \frac{0.2}{6.4/2} = \frac{0.2}{3.2} = \frac{1}{16} \)

\( \implies \omega_{VR} = 0.0625 \)
In simple words: Angular dispersion measures the difference in deviation for extreme colors, while dispersive power is the ratio of angular dispersion to mean deviation, both calculable from refractive indices and the prism angle.

🎯 Exam Tip: Ensure you use the correct refractive indices for violet and red light when calculating individual deviations and their difference for angular dispersion. The mean deviation is often taken as the deviation for yellow light or the average of extreme deviations.

Question 11. Refractive index of a flint glass varies from 1.60 to 1.66 for visible range. Radii of curvature of a thin convex lens are 10 cm and 15 cm. Calculate the chromatic aberration between extreme colours.
Answer:
Given the refractive indices for extreme colours. As, \( n_R < n_V \)
\( n_R = 1.60 \) and \( n_V = 1.66 \)
For convex lens,
\( R_1 = 10 \) cm and \( R_2 = -15 \) cm
From lens maker's equation:
\( \frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \)
For red colour,
\( \frac{1}{f_R} = (n_R - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \)

\( \implies \frac{1}{f_R} = (1.60 - 1) \left( \frac{1}{10} - \frac{1}{(-15)} \right) = 0.6 \left( \frac{1}{10} + \frac{1}{15} \right) = 0.6 \left( \frac{3+2}{30} \right) = 0.6 \times \frac{5}{30} = 0.6 \times \frac{1}{6} = 0.1 \)

\( \implies f_R = 10 \) cm
Similarly, for violet colour,
\( \frac{1}{f_V} = (n_V - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \)

\( \implies \frac{1}{f_V} = (1.66 - 1) \left( \frac{1}{10} - \frac{1}{(-15)} \right) = 0.66 \left( \frac{1}{10} + \frac{1}{15} \right) = 0.66 \left( \frac{5}{30} \right) = 0.66 \times \frac{1}{6} = 0.11 \)

\( \implies f_V = 11 \) cm
Longitudinal chromatic aberration
\( = f_V - f_R = 11 - 10 = 1 \) cm
In simple words: Chromatic aberration is the difference in focal lengths for different colors due to the lens material's dispersion. It can be calculated by finding the focal length for extreme colors (red and violet) using the lens maker's formula and then taking their difference.

🎯 Exam Tip: Remember that for a convex lens, violet light has a shorter focal length than red light because its refractive index is higher. Correctly apply sign conventions for \( R_1 \) and \( R_2 \) in the lens maker's formula to avoid errors.

Question 12. A person uses spectacles of 'number' 2.00 for reading. Determine the range of magnifying power (angular magnification) possible. It is a concave convex lens (n = 1.5) having curvature of one of its surfaces to be 10 cm. Estimate that of the other.
Answer:
For a single concavo-convex lens, the magnifying power will be same as that for simple microscope As, the number represents the power of the lens,
P = 2.00 D
\( \implies P = \frac{1}{f} = 2 \implies f = 0.5 \) m.
Range of magnifying power of a lens will be,
\( M_{min} = \frac{D}{f} = \frac{0.25}{0.5} = 0.5 \)
And \( M_{max} = 1 + \frac{D}{f} = 1 + 0.5 = 1.5 \)
Given that, \( n = 1.5 \), \( |R_1| = 10 \) cm
\( f = 0.5 \) m = 50 cm
From lens maker's equation,
\( \frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \)
For a concavo-convex lens, one surface is concave, one is convex. Let \( R_1 \) be convex, \( R_2 \) be concave.
\( \implies \frac{1}{50} = (1.5-1) \left( \frac{1}{10} - \frac{1}{R_2} \right) \)
\( \implies \frac{1}{50} = 0.5 \left( \frac{1}{10} - \frac{1}{R_2} \right) \)
\( \implies 0.02 = 0.5 \left( 0.1 - \frac{1}{R_2} \right) \)
\( \implies \frac{0.02}{0.5} = 0.1 - \frac{1}{R_2} \)
\( \implies 0.04 = 0.1 - \frac{1}{R_2} \)
\( \implies \frac{1}{R_2} = 0.1 - 0.04 = 0.06 \)
\( \implies R_2 = \frac{1}{0.06} = \frac{100}{6} = \frac{50}{3} \) cm
\( \implies R_2 \approx 16.67 \) cm
In simple words: The focal length of spectacles, derived from their power, determines the range of angular magnification. The unknown radius of curvature of a concavo-convex lens can then be found using the lens maker's formula.

🎯 Exam Tip: Remember that lens power (in dioptres) is the reciprocal of focal length (in meters). For concave-convex lenses, correctly assign positive and negative signs to the radii of curvature based on whether the surface is convex or concave relative to the incident light.

Question 13. Focal power of the eye lens of a compound microscope is 6 dioptre. The microscope is to be used for maximum magnifying power (angular magnification) of at least 12.5. The packing instructions demand that length of the microscope should be 25 cm. Determine minimum focal power of the objective. How much will its radius of curvature be if it is a biconvex lens of n = 1.5.
Answer:
Focal power of the eye lens,
\( P_e = \frac{1}{f_e} = 6 \) D

\( \implies f_e = \frac{1}{6} \) m \( = 0.1667 \) m \( = 16.67 \) cm
Now, as the magnifying power is maximum,
\( v_e = 25 \) cm (This is the least distance of distinct vision, D)
Also \( (M_e)_{max} = 1 + \frac{D}{f_e} = 1 + \frac{25}{16.67} \approx 2.5 \)
Given that,
\( M = m_o \times M_e = 12.5 \)
\( \implies m_o \times 2.5 = 12.5 \)

\( \implies m_o = \frac{12.5}{2.5} = 5 \) ........... (1)
From thin lens formula,
\( \frac{1}{f_e} = \frac{1}{v_e} - \frac{1}{u_e} \)

\( \implies \frac{1}{16.67} = \frac{1}{(-25)} - \frac{1}{u_e} \) (Using \( v_e = -D = -25 \) for virtual image at D)

\( \implies \frac{1}{u_e} = \frac{1}{(-25)} - \frac{1}{16.67} = - \frac{1}{25} - \frac{1}{16.67} = -0.04 - 0.06 = -0.1 \)

\( \implies u_e \approx -10 \) cm ............(ii)
Length of a compound microscope,
\( L = |v_o| + |u_e| \)

\( \implies 25 = |v_o| + 10 \)

\( \implies |v_o| = 15 \) cm
From \( m_o = \frac{v_o}{u_o} = 5 \) (from 1)

\( \implies |u_o| = \frac{|v_o|}{5} = \frac{15}{5} = 3 \) cm
From lens formula for objective,
\( \frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o} \)

\( \implies \frac{1}{f_o} = \frac{1}{15} - \frac{1}{(-3)} = \frac{1}{15} + \frac{1}{3} = \frac{1+5}{15} = \frac{6}{15} = \frac{2}{5} \)

\( \implies f_o = \frac{5}{2} = 2.5 \) cm \( = 0.025 \) m
Thus, focal power of objective,
\( P_o = \frac{1}{f_o(m)} = \frac{1}{0.025} = 40 \) D
Using lens maker's equation for a biconvex lens,
\( \frac{1}{f_o} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \)
For biconvex, \( R_1 = R \), \( R_2 = -R \).
\( \implies \frac{1}{0.025} = (1.5-1) \left( \frac{1}{R} - \frac{1}{(-R)} \right) = 0.5 \left( \frac{2}{R} \right) = \frac{1}{R} \)

\( \implies R = 0.025 \) m \( = 2.5 \) cm
In simple words: By using the given magnifying power and length constraints for a compound microscope, the focal lengths and object/image distances for both the objective and eyepiece can be calculated, leading to the determination of the objective lens's radii of curvature.

🎯 Exam Tip: Break down compound microscope problems into two lens calculations (eyepiece first, then objective). Remember that total magnification is the product of individual magnifications, and microscope length relates to the separation of the lenses. Always be mindful of sign conventions and unit conversions (cm to m for power calculations).

Can You Recall? (Textbook Page No 159)

What are laws of reflection and refraction?
Answer:
Laws of reflection:
a. Reflected ray lies in the plane formed by incident ray and the normal drawn at the point of incidence and the two rays are on either side of the normal.
b. Angles of incidence and reflection are equal (i = r).
Laws of refraction:
a. Refracted ray lies in the plane formed by incident ray and the normal drawn at the point of incidence; and the two rays are on either side of the normal.
b. Angle of incidence (\( \theta_1 \)) and angle of refraction (\( \theta_2 \)) are related by Snell's law, given by, \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \) where, \( n_1, n_2 \) = refractive indices of medium 1 and medium 2 respectively.
In simple words: The laws of reflection state that the angle of incidence equals the angle of reflection, and incident, reflected rays, and normal are in the same plane. The laws of refraction state that incident, refracted rays, and normal are in the same plane, and Snell's law (\( n_1 \sin \theta_1 = n_2 \sin \theta_2 \)) governs the relationship between incidence and refraction angles.

🎯 Exam Tip: Clearly state both parts of each law. For Snell's Law, ensure you correctly associate \( n_1 \) with \( \theta_1 \) and \( n_2 \) with \( \theta_2 \).

Can You Recall? (Textbook Page No. 159)

Question 1. What is refractive index?
Answer:
The ratio of velocity of light in vacuum to the velocity of light in a medium is called the refractive index of the medium.
In simple words: Refractive index is a measure of how much a medium slows down light, defined as the ratio of light speed in vacuum to its speed in that medium.

🎯 Exam Tip: Remember the definition of refractive index as a ratio of speeds. It's a dimensionless quantity and indicates optical density.

Question 2. What is total internal reflection?
Answer:
For angles of incidence larger than the critical angle, the angle of refraction is larger than 90°. Thus, all the incident light gets reflected back into the denser medium. This is called total internal reflection.
In simple words: Total internal reflection occurs when light traveling from a denser to a rarer medium hits the interface at an angle greater than the critical angle, causing all light to reflect back into the denser medium.

🎯 Exam Tip: Two critical conditions for Total Internal Reflection (TIR) are: light must travel from a denser to a rarer medium, and the angle of incidence must be greater than the critical angle.

Question 3. How does a rainbow form?
Answer:
1. The rainbow appears in the sky after a rainfall.
2. Water droplets present in the atmosphere act as small prism.
3. When sunlight enters these water droplets it gets refracted and dispersed.
4. This dispersed light gets totally reflected inside the droplet and again is refracted while coming out of the droplet.
5. As a combined effect of all these phenomena, the seven coloured rainbow is observed.
In simple words: Rainbows form when sunlight interacts with raindrops, which act like tiny prisms, causing light to refract, disperse into colors, totally internally reflect, and then refract again as it exits the droplet, creating the arc of colors.

🎯 Exam Tip: When explaining rainbow formation, mention the key optical phenomena: refraction, dispersion, and total internal reflection, occurring sequentially within raindrops.

Question 4. What is dispersion of light?
Answer:
Splitting of a white light into its constituent colours is known as dispersion of light.
In simple words: Dispersion of light is the phenomenon where white light separates into its component colors (like a rainbow) because each color travels at a slightly different speed through a medium, causing different refractive angles.

🎯 Exam Tip: The core idea of dispersion is the separation of white light into its spectrum due to the dependence of refractive index on wavelength (color).