RBSE Solutions Class 12 Physics Chapter 11 Ray Optics

RBSE Class 12 Physics Chapter 11 Multiple Choice Type Questions

 

Question 1. In spherical mirrors, we consider only paraxial rays for forming the images, because :
(a) They are easy to work for geometrical purpose
(b) They consist of mostly intense incident light
(c) They form point image of the point object
(d) They produce minimum dispersion
Answer: (c) They form point image of the point object
In simple words: We use only paraxial rays in spherical mirrors because they help create a clear, small image of a point. These rays cause the least blurring.

🎯 Exam Tip: Remember that paraxial rays are those close to the principal axis and make small angles with it, simplifying image formation calculations.

 

Question 2. An object is placed at 30 cm distance from a concave mirror of focal length 20 cm, then nature, position and magnification will be:
(a) Real, -2
(b) Virtual, 2
(c) Real, \( - \frac{1}{2} \)
(d) Virtual, \( \frac{1}{2} \)
Answer: (a) Real, -2
Given, \( f = -20 \text{ cm} \), \( u = -30 \text{ cm} \), \( v = ? \)
From equation, \( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \)
\( \frac{1}{v} + \frac{1}{-30} = \frac{1}{-20} \)
Now, we solve for \( \frac{1}{v} \):
\( \frac{1}{v} = \frac{1}{-20} - \frac{1}{-30} \)
\( \frac{1}{v} = - \frac{1}{20} + \frac{1}{30} \)
To add these fractions, find a common denominator, which is 60:
\( \frac{1}{v} = - \frac{3}{60} + \frac{2}{60} \)
\( \frac{1}{v} = \frac{-3+2}{60} \)
\( \frac{1}{v} = \frac{-1}{60} \)
\( v = -60 \text{ cm} \) (This means the image is real, formed on the same side as the object)
Next, calculate magnification \( m \):
\( m = - \frac{v}{u} \)
\( m = - \frac{-60}{-30} \)
\( m = -2 \)
Since the magnification is negative and greater than 1, the image is real, inverted, and enlarged.
In simple words: When an object is 30 cm from a concave mirror with a focal length of 20 cm, the image will form 60 cm away on the same side as the object. This image will be real and upside down, appearing twice as large as the object.

🎯 Exam Tip: Remember the sign conventions for mirrors: focal length of a concave mirror is negative, object distance is negative, and real images have negative image distances. A negative magnification means the image is inverted.

 

Question 3. The refractive index for infrared rays :
(a) is equal to ultraviolet rays
(b) is equal to red colour rays
(c) less than ultraviolet rays
(d) greater than ultraviolet rays
Answer: (b) is equal to red colour rays
In simple words: Infrared rays and red colour rays have similar properties. Because of this, the bending of light (refractive index) is the same for both when they pass through a material.

🎯 Exam Tip: Recall that infrared light is just beyond red light in the electromagnetic spectrum, and they behave similarly in terms of refraction in many materials.

 

Question 4. Total internal reflection occurs, when :
(a) Light travels from optically rarer medium to optically denser medium
(b) Light travels from optically denser medium to optically rarer medium
(c) Refractive indices of both medium are equal
(d) Refractive indices of both medium are different
Answer: (b) Light travels from optically denser medium to optically rarer medium
In simple words: Total internal reflection happens when light tries to move from a thicker material (like glass) to a thinner material (like air) and hits the surface at a very sharp angle. Instead of passing through, all the light bounces back inside the thicker material.

🎯 Exam Tip: For total internal reflection to occur, two conditions must be met: light must travel from a denser to a rarer medium, and the angle of incidence must be greater than the critical angle.

 

Question 6. A convex lens of power +6 D is in contact with a concave lens of power -4 D. What will be the focus length and nature of the combined lens?
(a) Concave, 25 cm
(b) Convex, 50 cm
(c) Concave, 20 cm
(d) Convex, 100 cm
Answer: (b) Convex, 50 cm
Given: Power of convex lens \( P_1 = +6 \text{ D} \)
Power of concave lens \( P_2 = -4 \text{ D} \)
The total power of the combined lens \( P \) is the sum of individual powers:
\( P = P_1 + P_2 \)
\( P = +6 \text{ D} + (-4 \text{ D}) \)
\( P = +2 \text{ D} \)
Since the combined power \( P \) is positive, the combined lens is convex.
The focal length \( f \) of the combined lens is given by \( f = \frac{1}{P} \). If power is in Dioptres, focal length in meters. To get it in cm, we use \( f = \frac{100}{P} \).
\( f = \frac{100}{+2} \text{ cm} \)
\( f = +50 \text{ cm} \)
So, the combined lens is convex with a focal length of 50 cm.
In simple words: When a strong convex lens and a weaker concave lens are put together, they act like a single convex lens. Its total strength is found by adding their powers, which gives +2 D. This means the combined lens is convex and has a bending distance (focal length) of 50 cm.

🎯 Exam Tip: Remember that for lenses in contact, the total power is the algebraic sum of individual powers. Positive power means a convex lens, and negative power means a concave lens.

 

Question 7. A ray passes through on equilateral triangle in such a way that its angle of incidence and angle of emergence are equal and this angle is 3/4 of angle of prism. Then angle of deviation will be :
(a) \( 45^\circ \)
(b) \( 70^\circ \)
(c) \( 39^\circ \)
(d) \( 30^\circ \)
Answer: (d) \( 30^\circ \)
For an equilateral triangle prism, the angle of prism \( A = 60^\circ \).
It is given that the angle of incidence \( i \) and angle of emergence \( e \) are equal:
\( i = e \)
Also, this angle is \( \frac{3}{4} \) of the angle of prism:
\( i = e = \frac{3}{4} A \)
Substitute \( A = 60^\circ \):
\( i = e = \frac{3}{4} \times 60^\circ \)
\( i = e = 45^\circ \)
The formula for the angle of deviation \( \delta \) in a prism is:
\( \delta = i + e - A \)
Substitute the values:
\( \delta = 45^\circ + 45^\circ - 60^\circ \)
\( \delta = 90^\circ - 60^\circ \)
\( \delta = 30^\circ \)
In simple words: For an equilateral prism, the prism angle is 60 degrees. If the light goes in and comes out at the same angle, and this angle is three-quarters of the prism angle (which is 45 degrees), then the total bending of the light (deviation) is 30 degrees.

🎯 Exam Tip: Remember the prism angle for an equilateral triangle is always \( 60^\circ \). The relation \( \delta = i + e - A \) is fundamental for prism problems.

 

Question 8. The image formed by objective lens of compound telescope will be :
(a) Virtual and bigger
(b) Virtual and small
(c) Real and point size
(d) Real and bigger
Answer: (d) Real and bigger
In simple words: The first lens in a compound telescope (called the objective lens) creates an image that is real and larger than the actual distant object. This image is then further magnified by the eyepiece.

🎯 Exam Tip: In a compound microscope, the objective produces a real, inverted, magnified image, whereas in a telescope, it produces a real, inverted, and diminished image of a distant object (which then acts as an object for the eyepiece).

 

Question 9. A covexo-convex lens of refractive index 1.47 is dipped in a liquid, then it behaves like a simple plane sheet of glass. It means, the refractive index of liquid is :
(a) Greater than the refractive index of glass
(b) Less than the refractive index of glass
(c) Equal to the refractive index of glass
(d) Less than one
Answer: (c) Equal to the refractive index of glass
In simple words: If a lens acts like a flat piece of glass when put in a liquid, it means the liquid bends light exactly the same way the lens material does. So, their refractive indices must be the same.

🎯 Exam Tip: A lens loses its converging or diverging power when placed in a medium with the same refractive index as its own material, behaving like a transparent sheet.

 

Question 10. The angle of minimum deviation of a prism will be equal to its angle of refractive index. If refractive index of prism is:
(a) 1
(b) 2
(c) \( \sqrt{2} \)
(d) 1.5
Answer: (b) 2
We are given that the angle of minimum deviation \( \delta_m \) is equal to the angle of the prism \( A \):
\( \delta_m = A \)
The formula for the refractive index \( n \) of a prism is:
\( n = \frac{\sin \left( \frac{A + \delta_m}{2} \right)}{\sin \left( \frac{A}{2} \right)} \)
Substitute \( \delta_m = A \) into the formula:
\( n = \frac{\sin \left( \frac{A + A}{2} \right)}{\sin \left( \frac{A}{2} \right)} \)
\( n = \frac{\sin \left( \frac{2A}{2} \right)}{\sin \left( \frac{A}{2} \right)} \)
\( n = \frac{\sin A}{\sin \left( \frac{A}{2} \right)} \)
Using the trigonometric identity \( \sin A = 2 \sin \left( \frac{A}{2} \right) \cos \left( \frac{A}{2} \right) \):
\( n = \frac{2 \sin \left( \frac{A}{2} \right) \cos \left( \frac{A}{2} \right)}{\sin \left( \frac{A}{2} \right)} \)
We can cancel \( \sin \left( \frac{A}{2} \right) \) from the numerator and denominator (assuming \( A \neq 0 \)):
\( n = 2 \cos \left( \frac{A}{2} \right) \)
We need to find the value of \( n \). Let's check the given options:
(i) If \( A \approx 0 \), then \( \cos \left( \frac{A}{2} \right) \approx 1 \), so \( n \approx 2 \times 1 = 2 \).
(ii) If \( A = 90^\circ \), then \( \frac{A}{2} = 45^\circ \).
\( n = 2 \cos 45^\circ \)
\( n = 2 \times \frac{1}{\sqrt{2}} \)
\( n = \frac{2}{\sqrt{2}} = \sqrt{2} \)
The value of \( n \) lies between \( \sqrt{2} \) and 2. Since 2 is one of the options and is a possible value when \( A \) is very small (or effectively \( \delta_m = A \) as a limiting case), it is the intended answer.
In simple words: When a prism's minimum bending angle is the same as its own top angle, its refractive index is found using a special formula. If the prism angle is very small, the refractive index will be close to 2. If the prism angle is 90 degrees, the refractive index is \( \sqrt{2} \). So, 2 is a possible value for the refractive index.

🎯 Exam Tip: Understand the relationship between the angle of minimum deviation, angle of prism, and refractive index. Remember the special case where \( \delta_m = A \) simplifies the refractive index formula to \( n = 2 \cos \left( \frac{A}{2} \right) \).

 

Question 11. A ray of light falling normally on a plane mirror, then angle of reflection will be :
(a) \( 90^\circ \)
(b) \( 180^\circ \)
(c) \( 0^\circ \)
(d) \( 45^\circ \)
Answer: (c) \( 0^\circ \)
In simple words: If a light ray hits a flat mirror straight on (at a 90-degree angle to the surface), it bounces directly back along the same path. This means its angle of reflection is zero.

🎯 Exam Tip: The law of reflection states that the angle of incidence equals the angle of reflection. When light falls normally, both angles are \( 0^\circ \) relative to the normal.

 

Question 12. If an object is placed at focus of a concave mirror, the image will be formed at :
(c) At O
(d) At \( \infty \)
Answer: (d) At \( \infty \)
In simple words: When an object sits exactly at the focal point of a concave mirror, the light rays bounce off the mirror and travel outwards as parallel lines. These parallel rays never meet, so the image appears to be formed infinitely far away.

🎯 Exam Tip: Recall the special ray diagrams for concave mirrors. Rays originating from the focus become parallel after reflection, hence forming an image at infinity.

 

Question 13. An observer is watching the Stars to be twinkle on Earth. The cause is :
(a) It is true that Stars do not emit light continuously
(b) Frequency absorption of light of Stars by atmosphere of their own
(c) Frequency absorption of light of Stars by atmosphere of Earth
(d) Increase and decrease of the refractive index in atmosphere of Earth
Answer: (d) Increase and decrease of the refractive index in atmosphere of Earth
In simple words: Stars appear to twinkle because the Earth's air is constantly moving and has different temperatures. This causes the light from the stars to bend and unbend slightly as it travels through the air, making the stars look like they are flickering.

🎯 Exam Tip: Twinkling of stars is a phenomenon caused by atmospheric refraction. Planets, being closer, do not twinkle because they appear as extended sources of light.

 

Question 14. If yellow light is refracted at angle of minimum deviation from prism, then :
(a) Angle of incidence and angle of emergence are equal
(b) The sum of the angle of incidence and angle of emergence is \( 90^\circ \)
(c) Angle of incidence is less than angle of emergence
(d) Angle of incidence is greater than angle of emergence
Answer: (a) Angle of incidence and angle of emergence are equal
In simple words: When light passes through a prism and bends the least (minimum deviation), the angle at which the light enters the prism is exactly the same as the angle at which it leaves the prism.

🎯 Exam Tip: A key condition for minimum deviation in a prism is that the angle of incidence (i) equals the angle of emergence (e), and the refracted ray inside the prism is parallel to its base.

 

Question 15. For a healthy eye, the least distance of distinct vision and maximum distance will be :
(a) 25 cm and 100 cm
(b) 25 cm and infinite
(c) 100 cm and infinite
(d) Zero and zero to infinite
Answer: (b) 25 cm and infinite
In simple words: For eyes that can see perfectly, the closest clear vision is about 25 cm away. The farthest clear vision for healthy eyes goes on forever, or to infinity.

🎯 Exam Tip: Remember these standard values for a healthy human eye: the near point is 25 cm (known as the least distance of distinct vision), and the far point is infinity.

 

Question 17. Apparent image of size greater than object may be formed by :
(a) Convex mirror
(b) Concave mirror
(c) Plane mirror
(d) Concave lens
Answer: (b) Concave mirror
In simple words: A concave mirror can make an object look bigger than it actually is. This happens when the object is placed close to the mirror, inside its focal point.

🎯 Exam Tip: Concave mirrors are used for shaving or makeup because they can produce magnified virtual images when the object is placed between the pole and the focal point.

 

Question 18. In compound microscope the final image is formed :
(a) Real and erect
(b) Virtual and inverted
(c) Virtual and erect
(d) Real and inverted
Answer: (b) Virtual and inverted
In simple words: In a compound microscope, the final image you see is not real; it's a virtual image, and it's also upside down compared to the original object.

🎯 Exam Tip: Remember that the final image in a compound microscope is virtual, inverted, and highly magnified. The objective lens forms a real, inverted image, which the eyepiece then magnifies into a virtual, inverted final image.

 

Question 19. In reflecting type telescope, the objective used is :
(a) Convex lens
(b) Convex mirror
(c) Concave mirror
(d) Concave lens
Answer: (c) Concave mirror
In simple words: A reflecting telescope uses a large curved mirror that is concave to collect and focus light from distant objects. This mirror is its main part.

🎯 Exam Tip: Reflecting telescopes use mirrors (usually concave) as objectives to avoid chromatic aberration and to build larger apertures for better light gathering.

 

Question 20. The power of the objective lens and eyepiece of an astronomical telescope are 5 D and 20 D. They form image at infinity. What will be the magnifying power of telescope?
(a) 4
(b) 2
(c) 100
(d) 0.25
Answer: (a) 4
Given: Power of objective lens \( P_o = +5 \text{ D} \)
Power of eyepiece \( P_e = +20 \text{ D} \)
Focal length of objective lens \( f_o = \frac{100}{P_o} = \frac{100}{5} = +20 \text{ cm} \)
Focal length of eyepiece \( f_e = \frac{100}{P_e} = \frac{100}{20} = +5 \text{ cm} \)
For an astronomical telescope forming an image at infinity, the magnifying power \( m \) is given by:
\( m = - \frac{f_o}{f_e} \)
The negative sign indicates the final image is inverted. For calculating the magnitude:
\( |m| = \frac{f_o}{f_e} \)
\( |m| = \frac{20 \text{ cm}}{5 \text{ cm}} \)
\( |m| = 4 \)
In simple words: To find how much a telescope magnifies, we first find the bending distance (focal length) of its two lenses. The objective lens has a focal length of 20 cm, and the eyepiece has 5 cm. Dividing these gives a magnifying power of 4.

🎯 Exam Tip: Remember that magnifying power for a telescope in normal adjustment (image at infinity) is the ratio of the focal length of the objective to that of the eyepiece (\( M = f_o / f_e \)). Ensure units are consistent (convert power to focal length in cm first).

 

Question 21. The power of convex lens is :
(a) Negative
(b) Positive
(c) Zero
(d) Imaginary
Answer: (b) Positive
In simple words: A convex lens focuses light rays inwards. Because it converges light, its power, which describes how much it bends light, is always considered positive.

🎯 Exam Tip: Convex lenses are converging lenses, meaning they have a real focus and a positive focal length, which in turn means they have positive power. Concave lenses are diverging and have negative power.

RBSE Class 12 Physics Chapter 11 Very Short Answer Type Questions

 

Question 1. What is the focal length of a plane mirror?
Answer: The focal length of a plane mirror is considered to be infinite.
In simple words: A flat mirror's focal length is endless, meaning it doesn't bring light to a specific focus or spread it out.

🎯 Exam Tip: Remember that a plane mirror forms a virtual image at the same distance behind the mirror as the object is in front. Its flat surface implies an infinite radius of curvature, leading to an infinite focal length.

 

Question 3. What is the cause of refraction?
Answer: The main reason for refraction is that the speed of light changes when it travels from one medium to another. Different materials slow light down by different amounts, causing it to bend.
In simple words: Light bends (refracts) because its speed changes when it goes from one material to another.

🎯 Exam Tip: Relate the change in speed of light to the change in wavelength when moving between media, which causes the bending of light. Frequency remains constant.

 

Question 4. What is the cause of mirage in desert area?
Answer: The cause of a mirage in a desert area is total internal reflection. This happens due to the extreme heating of the sand, which makes the air layers near the ground much hotter and less dense than the cooler air layers above. This creates a gradient in the refractive index of the atmosphere, leading to total internal reflection.
In simple words: Mirages in deserts happen because of total internal reflection. Hot air near the ground bends light so much that it reflects upwards, making it look like there's water.

🎯 Exam Tip: Emphasize that mirages are an optical illusion caused by the bending of light in layers of air with different temperatures and thus different refractive indices.

 

Question 5. For equal angle of incidence, the angle of refraction for three medium A B and C are \( 15^\circ \), \( 25^\circ \) and \( 35^\circ \). In which medium the velocity of light will be minimum?
Answer: The refractive index \( \mu \) of a medium is related to the angle of incidence \( i \) and angle of refraction \( r \) by Snell's Law: \( \mu = \frac{\sin i}{\sin r} \).
Given that the angle of incidence \( i \) is constant for all three media, and \( \sin i \) is constant.
So, \( \mu \propto \frac{1}{\sin r} \). This means that a smaller angle of refraction \( r \) results in a larger refractive index \( \mu \).
The angles of refraction are given as \( r_A = 15^\circ \), \( r_B = 25^\circ \), and \( r_C = 35^\circ \).
From these values, we can see that \( \sin r_A < \sin r_B < \sin r_C \).
Therefore, the refractive indices will follow this order: \( \mu_A > \mu_B > \mu_C \).
The velocity of light \( v \) in a medium is related to the refractive index \( \mu \) by \( v = \frac{c}{\mu} \), where \( c \) is the speed of light in vacuum. This means that a larger refractive index results in a smaller velocity of light.
Since \( \mu_A \) is the largest, the velocity of light will be minimum in medium A.
In simple words: Light bends least in medium A (15 degrees), which means medium A has the highest refractive index. When a material has a higher refractive index, light travels slower in it. So, light's speed will be slowest in medium A.

🎯 Exam Tip: Remember that a smaller angle of refraction implies a denser medium, which means a higher refractive index and a slower speed of light. Conversely, a larger angle of refraction implies a rarer medium, a lower refractive index, and a faster speed of light.

 

Question 6. Name the principle on which optical fibre works.
Answer: Optical fibre works on the principle of total internal reflection.
In simple words: Optical fibers use total internal reflection to carry light signals over long distances without much loss.

🎯 Exam Tip: Optical fibers are crucial in communication and medical imaging. Be sure to understand how the critical angle and refractive index difference between the core and cladding enable total internal reflection.

 

Question 7. At the position of minimum deviation what is relation between angle of incidence and angle of emergence?
Answer: At the position of minimum deviation, the angle of incidence is equal to the angle of emergence. This means that \( i = e \).
In simple words: When a prism bends light the least, the angle at which the light goes in is exactly the same as the angle at which it comes out.

🎯 Exam Tip: For minimum deviation, the path of the light ray inside the prism is symmetrical, making the angle of incidence equal to the angle of emergence, and the refracted ray parallel to the base of the prism.

 

Question 9. Sun looks reddish at sunset or sunrise. Why?
Answer: During sunset or sunrise, the sun appears reddish because of the scattering of light. At these times, sunlight has to travel a much longer distance through the Earth's atmosphere. Most of the blue light and other shorter wavelengths are scattered away by the air particles, leaving mainly the longer wavelength red light to reach our eyes directly. This phenomenon follows Rayleigh's scattering law, where the intensity of scattered light \( I_s \propto \frac{1}{\lambda^4} \). Therefore, shorter wavelengths (like blue) scatter much more efficiently than longer wavelengths (like red).
In simple words: The sun looks red at sunrise and sunset because sunlight travels through more air. Blue light gets scattered away more easily by the air, so mostly red light makes it directly to our eyes.

🎯 Exam Tip: Remember Rayleigh scattering: shorter wavelengths scatter more effectively. At sunrise/sunset, the increased atmospheric path length scatters away blue light, allowing red light to dominate.

 

Question 10. What is the cause of rainbow?
Answer: The formation of a rainbow is caused by the dispersion of sunlight by tiny water drops present in the atmosphere. When sunlight enters a raindrop, it first refracts and splits into its different colors (dispersion). Then, it undergoes total internal reflection inside the raindrop, and finally, it refracts again as it exits the raindrop, further separating the colors into a beautiful arc.
In simple words: A rainbow happens when sunlight passes through raindrops. The raindrops act like tiny prisms, splitting the white light into all its different colors and reflecting it back to our eyes.

🎯 Exam Tip: Key phenomena for rainbow formation are dispersion, total internal reflection, and refraction. All colors of visible light are involved, with different angles for red and violet light.

 

Question 11. What is myopia? How can it be removed?
Answer: Myopia, also known as nearsightedness, is a vision defect where a person can see nearby objects clearly but cannot see distant objects distinctly. This happens because the eye lens focuses light from distant objects in front of the retina. It can be corrected by using a concave lens of appropriate focal length. This lens diverges the light rays slightly before they enter the eye, allowing them to focus correctly on the retina.
In simple words: Myopia means you can see close things well but far things are blurry. It's fixed by wearing glasses with concave lenses.

🎯 Exam Tip: Clearly distinguish myopia (nearsightedness) from hypermetropia (farsightedness) by symptoms and the type of corrective lens (concave for myopia, convex for hypermetropia).

 

Question 12. Name the factor on which the scattering of light depends.
Answer: The scattering of light depends mainly on its wavelength. According to Rayleigh's scattering law, the intensity of scattered light is inversely proportional to the fourth power of its wavelength (\( I_s \propto \frac{1}{\lambda^4} \)). This means shorter wavelengths (like blue light) scatter much more than longer wavelengths (like red light).
In simple words: How much light scatters depends on its color, or wavelength. Shorter wavelengths (like blue) scatter a lot, while longer wavelengths (like red) scatter less.

🎯 Exam Tip: Mentioning Rayleigh's scattering law and its dependence on \( \lambda^4 \) is key to a complete answer, explaining phenomena like the blue sky and red sunsets.

 

Question 13. What type of lens is used in simple microscope?
Answer: A simple microscope uses a convex lens of short focal length. This lens is used to produce a magnified virtual image of a small object placed just inside its focal point.
In simple words: A simple microscope uses one convex lens with a short focal length to make small things look bigger.

🎯 Exam Tip: Remember that a convex lens acts as a simple microscope when the object is placed between the optical center and the focal point, creating a magnified, virtual, and erect image.

 

Question 2. Write uses of spherical mirrors.
Answer: Spherical mirrors are of two main types: concave mirrors and convex mirrors.
(a) Uses of Concave Mirrors:
1. Concave mirrors are used as shaving or makeup mirrors. When an object (like a face) is held close to the mirror, it forms a magnified and erect (upright) image, which helps in better viewing.
2. They are used as reflectors in headlights of cars, torches, and searchlights. The light source is placed at the focus of the concave mirror, so after reflection, the light rays travel as a parallel, intense beam over a large distance.
3. They are also used by dentists to see larger images of the teeth.
(b) Uses of Convex Mirrors:
1. Convex mirrors are commonly used as rear-view mirrors in automobiles. They always form a small and erect (upright) image, and importantly, they provide a much larger field of view compared to a plane mirror of the same size. This helps the driver see a wider area behind the vehicle.
2. They are used as security mirrors in shops, allowing a wide view of the store to deter theft.
In simple words: Concave mirrors are used for things like shaving or car headlights because they can make things look bigger or focus light into a strong beam. Convex mirrors are used as car side mirrors or security mirrors in shops because they show a wider area, making things look smaller but giving a broader view.

🎯 Exam Tip: For spherical mirrors, remember the different image formations for concave (real/virtual, magnified/diminished depending on object position) and convex (always virtual, erect, diminished), which dictates their practical applications.

 

Question 3. What is relation between focal length and radius of curvature for a mirror? Derive the relation.
Answer: The relationship between the focal length (\( f \)) and the radius of curvature (\( R \)) for a spherical mirror is \( f = \frac{R}{2} \). This means the focal length is half of the radius of curvature.
**Derivation (for Convex Mirror):**
Let \( M_1 M_2 \) be a convex mirror with pole \( P \) and center of curvature \( C \). Let \( F \) be the principal focus.
Consider an incident ray \( OA \) parallel to the principal axis. After reflection, this ray \( AS \) appears to come from the focus \( F \).
Draw a normal \( AN \) to the spherical surface at point \( A \). This normal passes through the center of curvature \( C \).
According to the law of reflection, the angle of incidence \( i \) equals the angle of reflection \( r \):
\( \angle OAN = \angle NAS = i \)
Since \( OA \) is parallel to the principal axis \( PC \), and \( AN \) is the transversal:
\( \angle OAN = \angle ANC \) (Alternate interior angles).
So, \( i = \angle ANC \).
Also, \( \angle NAS = \angle ACF \) (Vertically opposite angles).
Since \( \angle OAN = \angle NAS \), we have \( \angle ANC = \angle ACF \).
In triangle \( AFC \), because \( \angle ACF = \angle FAC \) (since \( \angle ANC = \angle ACF \)), the triangle \( AFC \) is isosceles with \( AF = FC \).
If the aperture of the mirror is small, point \( A \) is very close to point \( P \). Therefore, \( AF \) can be approximated as \( PF \).
So, \( PF = FC \).
The radius of curvature \( R \) is the distance \( PC \). From the figure, \( PC = PF + FC \).
Substitute \( PF = FC \):
\( R = PF + PF \)
\( R = 2 PF \)
The focal length \( f \) is defined as \( PF \). So, \( f = PF \).
Therefore, \( R = 2f \)
This gives us the relation: \( f = \frac{R}{2} \).
This derivation is also true for a concave mirror.
In simple words: The focal length of a curved mirror is always half of its radius of curvature. We can prove this by tracing light rays: a ray coming parallel to the mirror's center line bounces off and seems to come from the focus. By using geometry and the law of reflection, we find that the distance to the focus is exactly half the distance to the center of the curve.

🎯 Exam Tip: This derivation is crucial. Clearly state the law of reflection and geometric properties like alternate interior angles. Remember the small aperture approximation where \( AF \approx PF \).

 

Question 4.
(i) Sun looks reddish at sunset and sunrise. Why? Explain.
(ii) For which colour the refractive index of the prism is maximum and minimum?
Answer:
(i) During sunrise and sunset, the Sun appears reddish because of the scattering of light. At these times, sunlight travels a much longer distance through the Earth's atmosphere. Most of the blue light and other shorter wavelengths are scattered away by the air particles, following Rayleigh's scattering law (\( I_s \propto \frac{1}{\lambda^4} \)). This means that longer wavelength red light is less scattered and reaches our eyes directly, making the Sun appear reddish. This extra path length in the atmosphere effectively filters out the blue light.
(ii) The refractive index of the prism is maximum for violet colour and minimum for red colour. This is because violet light has the shortest wavelength in the visible spectrum and bends the most (largest refractive index), while red light has the longest wavelength and bends the least (smallest refractive index). Therefore, \( \mu_{violet} > \mu_{red} \).
In simple words:
(i) The sun looks red at sunrise and sunset because its light travels through a lot of air. The air scatters blue light away, so only red light makes it to our eyes.
(ii) For a prism, violet light bends the most, so it has the highest refractive index. Red light bends the least, so it has the lowest refractive index.

🎯 Exam Tip: Connect part (i) to Rayleigh scattering and the atmospheric path length. For part (ii), remember that refractive index varies with wavelength (dispersion), with shorter wavelengths (violet) generally experiencing higher refractive indices and longer wavelengths (red) experiencing lower refractive indices.

 

Question 5.
(i) What is relation between critical angle and refractive index for a substance?
(ii) Does critical angle depend on colour of light?
Answer:
(i) The relation between the critical angle (\( i_c \)) and the refractive index (\( \mu \)) for a substance when light travels from a denser medium to a rarer medium is given by:
\( \mu = \frac{1}{\sin i_c} \)
Here, \( \mu \) is the refractive index of the denser medium with respect to the rarer medium.
Alternatively, \( \sin i_c = \frac{1}{\mu} \).
(ii) Yes, the critical angle depends on the colour of light. This is because the refractive index \( \mu \) of a medium varies with the wavelength (colour) of light (a phenomenon called dispersion). Since \( \sin i_c = \frac{1}{\mu} \), if \( \mu \) changes with colour, then \( i_c \) must also change with colour. For instance, violet light (shorter wavelength) has a higher refractive index than red light (longer wavelength), meaning the critical angle for violet light will be smaller than for red light.
In simple words:
(i) The critical angle and refractive index are linked by a simple formula: \( \text{refractive index} = 1 / \sin(\text{critical angle}) \).
(ii) Yes, the critical angle changes for different colors of light. This is because each color bends differently, meaning each has a slightly different refractive index in the material.

🎯 Exam Tip: Remember that critical angle is crucial for total internal reflection. Its dependence on the refractive index and thus on the wavelength (color) of light is a direct consequence of dispersion.

 

Question 6. On what factors the focal length of the lens depend?
Answer: The focal length (\( f \)) of a lens depends on several factors, as described by the Lens Maker's Formula: \( \frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \).
Based on this formula, the focal length depends on the following factors:
1. **Radii of Curvature of the Lens Surfaces (\( R_1 \) and \( R_2 \)):** The curvature of the two surfaces of the lens determines how much the light rays bend. Sharper curves (smaller radii) generally lead to shorter focal lengths.
2. **Refractive Index of the Lens Material (\( n \)):** The refractive index of the material from which the lens is made. A higher refractive index means light bends more, resulting in a shorter focal length.
3. **Refractive Index of the Surrounding Medium:** Although not explicitly shown as a separate variable \( n_{medium} \) in the simplified formula, the term \( n \) in \((n-1)\) is actually \( \frac{n_{lens}}{n_{medium}} \), the relative refractive index of the lens material with respect to the surrounding medium. If the lens is placed in a medium with a different refractive index (e.g., water instead of air), its focal length will change.
In simple words: The focal length of a lens depends on three main things: how curved its surfaces are, what material the lens is made of (how much it bends light), and what material it is placed in (like air or water).

🎯 Exam Tip: The Lens Maker's Formula is key here. Make sure to explain each variable's impact on focal length. Remember the relative refractive index if the lens is not in air.

 

Question 7. How can you increase magnifying power of compound microscope?
Answer: To increase the magnifying power of a compound microscope, the following adjustments can be made:
1. **Decrease the focal length of the objective lens (\( f_o \)):** A shorter focal length for the objective lens means it produces a larger real image, which is then further magnified by the eyepiece.
2. **Decrease the focal length of the eyepiece (\( f_e \)):** A shorter focal length for the eyepiece also contributes to higher overall magnification.
3. **Increase the length of the microscope tube (distance between objective and eyepiece):** Increasing this distance generally allows for a larger intermediate image from the objective, which the eyepiece then magnifies. However, this is usually designed within limits.
Therefore, for high magnifying power in a compound microscope, both \( f_o \) and \( f_e \) should be small. Additionally, to enlarge the visible region, \( f_o \) should be less than \( f_e \).
In simple words: To make a compound microscope magnify more, you need to use objective and eyepiece lenses that have very short focal lengths. Making the tube longer can also help increase the magnification.

🎯 Exam Tip: Remember the formula for magnifying power of a compound microscope \( M = - \frac{L}{f_o} \left( 1 + \frac{D}{f_e} \right) \) (for image at near point) or \( M = - \frac{L}{f_o} \frac{D}{f_e} \) (for image at infinity). This clearly shows the inverse relationship with \( f_o \) and \( f_e \).

 

Question 8. What do you mean by scattering of light? Write its use in daily life.
Answer: Scattering of light refers to the phenomenon where light rays are deflected in various directions when they strike particles or molecules in a medium. The amount of scattering depends on the size of the particles and the wavelength of the light.
Uses in daily life:
1. **Blue colour of the sky:** The sky appears blue because short wavelength blue light is scattered more by the small particles and molecules in the Earth's atmosphere than longer wavelength colours.
2. **Reddishness at sunset and sunrise:** As explained before, during sunrise and sunset, the light travels a longer path through the atmosphere, causing most of the blue light to scatter away, leaving the longer wavelength red light to reach our eyes.
3. **Clouds appear white:** Clouds are made of larger water droplets and ice crystals, which are much larger than the wavelength of visible light. These larger particles scatter all wavelengths of light almost equally, making clouds appear white.
4. **Danger signals are red:** Red light is used for danger signals (like stop signs, traffic lights, and tail lights) because it has the longest wavelength among visible colours and scatters the least. This allows red light to travel the farthest through fog, smoke, or dust, making it visible from a greater distance.
In simple words: Scattering of light is when light hits tiny bits in the air and spreads out in different directions. This is why the sky looks blue, the sun looks red at morning and evening, clouds look white, and danger signals are red because red light travels far without scattering much.

🎯 Exam Tip: Explain scattering as redirection of light. When giving examples, clearly link each phenomenon (blue sky, red sunsets, white clouds, red danger signals) to the principle of wavelength-dependent scattering (Rayleigh scattering for small particles, equal scattering for large particles).

 

Question 9. Define power of the lens. Write its unit. For two thin lenses in combination, derive the relation \( \frac{1}{F}=\frac{1}{f_{1}}+\frac{1}{f_{2}} \)
Answer: The power of a lens describes how much it can bend light rays that come parallel to its main axis. It is measured by how much the lens deviates these rays when they hit it at a unit distance from the principal axis. The unit of power of a lens is Dioptre (D).
Combination of Thin Lenses:Consider two thin lenses, \( L_1 \) and \( L_2 \), with focal lengths \( f_1 \) and \( f_2 \) respectively, placed in contact with each other. A point object \( O \) is placed at a distance \( u \) in front of this combination. The final image is formed at \( I \).
The first lens, \( L_1 \), forms an intermediate image \( I' \) of the object \( O \). For lens \( L_1 \): \( \frac{1}{v'} - \frac{1}{u} = \frac{1}{f_1} \) ... (Equation 1)
This intermediate image \( I' \) then acts as a virtual object for the second lens, \( L_2 \). The second lens forms the final image at \( I \). For lens \( L_2 \): \( \frac{1}{v} - \frac{1}{v'} = \frac{1}{f_2} \) ... (Equation 2)
Now, we add Equation (1) and Equation (2): \( (\frac{1}{v'} - \frac{1}{u}) + (\frac{1}{v} - \frac{1}{v'}) = \frac{1}{f_1} + \frac{1}{f_2} \)
\( \implies \frac{1}{v} - \frac{1}{u} = \frac{1}{f_1} + \frac{1}{f_2} \) ... (Equation 3)
If these two lenses are considered as a single combined lens with an equivalent focal length \( F \), then for the object \( O \) at distance \( u \) and final image \( I \) at distance \( v \), the lens formula is: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{F} \) ... (Equation 4)
By comparing Equation (3) and Equation (4), we get: \( \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \)
This is the relation for the equivalent focal length of two thin lenses kept in contact. If there are more lenses, their powers simply add up.
In simple words: Lens power is how much it bends light. Its unit is Dioptre. When two thin lenses touch, their combined power is found by adding the individual powers. This makes the math for the new focal length simple to calculate.

🎯 Exam Tip: Remember that for lenses in contact, their powers add directly, \( P = P_1 + P_2 \), and since \( P = \frac{1}{F} \), this leads to the formula for combined focal length. Pay attention to sign conventions for focal lengths of convex (+) and concave (-) lenses.

 

RBSE Class 12 Physics Chapter 11 Long Answer Type Questions

 

Question 1. Define spherical mirror. Derive the formula connecting object distance, image distance and focal length for the mirror.
Answer: A spherical mirror is a reflective surface that is part of a hollow sphere. When light hits this polished surface, it reflects. There are two main types of spherical mirrors:
(i) Concave Mirror: This is formed when the outer, convex surface of a spherical glass plate is polished, making the inner, concave surface reflective. It focuses light.
(ii) Convex Mirror: This is formed when the inner, concave surface of a spherical glass plate is polished, making the outer, convex surface reflective. It spreads out light.
Derivation of Mirror Formula (for a convex mirror):Let's consider a convex mirror with focal length \( f \) and radius of curvature \( R \). A point object \( A \) is placed on the principal axis.
Figure 11.11 (not shown) illustrates a convex mirror with an object AB, image A'B', principal axis, and incident/reflected rays.
According to the law of reflection, the angle of incidence \( i \) is equal to the angle of reflection \( r \).
From the geometry of similar triangles (ΔABP and ΔA'B'P, where P is the pole of the mirror): \( \frac{A'B'}{AB} = \frac{PA'}{PA} \) ... (Equation 1)
Also, from similar triangles (ΔANC and ΔA'C'N where C is center of curvature): \( \frac{A'B'}{AB} = \frac{A'C}{AC} \) ... (Equation 2)
Comparing (1) and (2): \( \frac{PA'}{PA} = \frac{A'C}{AC} \)
We can write \( A'C = PC - PA' \) and \( AC = PA + PC \).
So, \( \frac{PA'}{PA} = \frac{PC - PA'}{PA + PC} \) ... (Equation 3)
Now, using the sign convention for a convex mirror: Object distance \( PA = -u \) (object is in front of the mirror, to the left) Image distance \( PA' = +v \) (image is formed behind the mirror, to the right) Radius of curvature \( PC = +R \) (center of curvature is behind the mirror, to the right)
Substitute these values into Equation (3): \( \frac{v}{-u} = \frac{R - v}{-u + R} \)
\( \implies v(-u + R) = -u(R - v) \)
\( \implies -uv + vR = -uR + uv \)
\( \implies vR + uR = 2uv \)
Divide the entire equation by \( uvR \): \( \frac{vR}{uvR} + \frac{uR}{uvR} = \frac{2uv}{uvR} \)
\( \implies \frac{1}{u} + \frac{1}{v} = \frac{2}{R} \)
Since for spherical mirrors, the focal length \( f \) is half the radius of curvature, \( R = 2f \) or \( \frac{2}{R} = \frac{1}{f} \).
Therefore, the mirror formula is: \( \frac{1}{u} + \frac{1}{v} = \frac{1}{f} \)
This formula connects the object distance (\( u \)), image distance (\( v \)), and focal length (\( f \)) for a spherical mirror. This formula is true for both concave and convex mirrors.
The magnification (\( m \)) of the image is the ratio of the image height (\( h' \)) to the object height (\( h \)), and also \( m = -\frac{v}{u} \).
In simple words: A spherical mirror is a curved shiny surface. There are two types: concave (inward curve) and convex (outward curve). The mirror formula, \( \frac{1}{u} + \frac{1}{v} = \frac{1}{f} \), tells us where the image will form based on where the object is and how curved the mirror is. This formula helps us understand how mirrors make images.

🎯 Exam Tip: Always use the correct sign conventions for u, v, f, and R when applying the mirror formula. For a concave mirror, f and R are negative. For a convex mirror, f and R are positive. Magnification indicates if the image is real/virtual and erect/inverted.

 

Question 3. Write the types of lenses. Using the diagram, derive the relation between object distance, image distance and focal length a lens.
Answer: A lens is a transparent optical element that focuses or diverges light rays. It is usually made from glass or plastic.
There are primarily two types of lenses:
1. Convex Lens (Converging Lens): These lenses are thicker in the middle and thinner at the edges. They converge (bring together) parallel rays of light to a single point called the principal focus. They can form both real and virtual images.
Types of Convex Lenses: (i) Double convex or biconvex lens: Both surfaces are convex. (ii) Plano-convex lens: One surface is plane and the other is convex. (iii) Concavo-convex lens: One surface is concave and the other is convex.
2. Concave Lens (Diverging Lens): These lenses are thinner in the middle and thicker at the edges. They diverge (spread out) parallel rays of light. These rays appear to originate from a single point on the principal axis after passing through the lens. They typically form virtual, erect, and diminished images.
Types of Concave Lenses: (i) Double concave or biconcave lens: Both surfaces are concave. (ii) Plano-concave lens: One surface is plane and the other is concave. (iii) Convexo-concave lens: One surface is convex and the other is concave.
Figure 11.27 and Figure 11.28 (not shown) illustrate the different types of convex and concave lenses. Figure 11.29 (not shown) illustrates the converging action of a convex lens and diverging action of a concave lens.
Lens Maker's Formula and Lens Formula Derivation (for a thin lens in air):Consider a thin lens with refractive index \( n \) placed in air. The radii of curvature of its two surfaces are \( R_1 \) and \( R_2 \). An object \( O \) is placed on the principal axis.
Figure 11.30 (not shown) illustrates a thin lens with radii of curvature \( R_1 \) and \( R_2 \), showing an object O, intermediate image I', and final image I.
For refraction at the first surface (curved surface facing the object, radius \( R_1 \)): The formula for refraction at a spherical surface is: \( \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \) Here, \( n_1 = 1 \) (air), \( n_2 = n \) (lens material). Let \( I' \) be the image formed by the first surface at distance \( v' \). So, \( \frac{n}{v'} - \frac{1}{u} = \frac{n - 1}{R_1} \) ... (Equation 1)
The image \( I' \) formed by the first surface acts as a virtual object for the second surface. The final image \( I \) is formed at distance \( v \). For the second surface, light travels from the lens material (refractive index \( n \)) to air (refractive index 1). So, \( n_1 = n \) (lens material), \( n_2 = 1 \) (air). The object distance for the second surface is \( v' \) (assuming the lens is thin, so thickness \( t \approx 0 \)). The radius of curvature for the second surface is \( R_2 \). Since the convex side of \( R_2 \) would be towards the right, if this is a double convex lens, \( R_2 \) will be negative when light passes through it from left to right. So, \( \frac{1}{v} - \frac{n}{v'} = \frac{1 - n}{R_2} \)
\( \implies \frac{1}{v} - \frac{n}{v'} = -\frac{(n - 1)}{R_2} \) ... (Equation 2)
Add Equation (1) and Equation (2): \( (\frac{n}{v'} - \frac{1}{u}) + (\frac{1}{v} - \frac{n}{v'}) = \frac{n - 1}{R_1} - \frac{n - 1}{R_2} \)
\( \implies \frac{1}{v} - \frac{1}{u} = (n - 1) (\frac{1}{R_1} - \frac{1}{R_2}) \) ... (Equation 3)
For a lens, if the object is at infinity (\( u = \infty \)), the image is formed at the principal focus (\( v = f \)). Substitute \( u = \infty \) and \( v = f \) into Equation (3): \( \frac{1}{f} - \frac{1}{\infty} = (n - 1) (\frac{1}{R_1} - \frac{1}{R_2}) \)
\( \implies \frac{1}{f} = (n - 1) (\frac{1}{R_1} - \frac{1}{R_2}) \)
This is the Lens Maker's Formula.
Comparing this with Equation (3), we get the Lens Formula: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \)
This formula relates the object distance (\( u \)), image distance (\( v \)), and focal length (\( f \)) for a thin lens.
In simple words: Lenses are clear pieces of material that bend light. They come in two main types: convex lenses (thicker in the middle, focus light) and concave lenses (thinner in the middle, spread light). The lens formula \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \) helps us find where an image will appear when light passes through a lens, based on where the object is and the lens's focal length.

🎯 Exam Tip: Remember that convex lenses have positive focal lengths and concave lenses have negative focal lengths. Always use the Cartesian sign convention consistently when applying the lens formula, especially for object and image distances.

 

Sign Convention for Spherical Surfaces

Answer: When dealing with refraction through spherical surfaces, we follow specific rules for measuring distances, similar to those for plane surfaces:
1. Light rays always fall on the spherical surface from the left side.
2. All distances are measured from the pole of the spherical surface along its principal axis.
3. Distances measured in the same direction as the incident light ray are taken as positive.
4. Distances measured in the opposite direction to the incident light ray are taken as negative.
5. The heights of objects and images are taken as positive if they are above the principal axis and negative if they are below it.
In simple words: When light bends through curved surfaces, we use rules to measure distances. Light always comes from the left. Distances measured in the same direction as light are positive, and those in the opposite direction are negative. Heights above the central line are positive, and below it are negative.

🎯 Exam Tip: Mastering sign conventions is crucial for solving problems in ray optics. A small error in assigning a sign can lead to an incorrect final answer.

 

Formula for Refraction of Convex Spherical Surface

Answer: Let's derive the formula for refraction at a convex spherical surface.
Consider a convex spherical surface AB (not shown, but Figure 11.26 in the source illustrates this with object O, image I, and point M on the surface). On its left side is a rarer medium (1) and on the right is a denser medium (2). Let P be the pole and C be the center of curvature.
According to Snell's law, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant, and equal to the refractive index of medium 2 with respect to medium 1: \( _1n_2 = \frac{\sin i}{\sin r} \)
If the angles \( i \) and \( r \) are very small (paraxial rays), then \( \sin i \approx i \) and \( \sin r \approx r \).
So, \( n = \frac{i}{r} \implies i = nr \) ... (Equation 1)
Now, consider the geometry. Let \( \alpha \), \( \beta \), and \( \gamma \) be the angles made by the incident ray, refracted ray, and principal axis with the optical axis respectively (refer to Figure 11.26 in source). From ΔMOC, \( i = \alpha + \gamma \) ... (Equation 2) From ΔMIC, \( \gamma = r + \beta \implies r = \gamma - \beta \) ... (Equation 3)
Substitute \( i \) and \( r \) from (2) and (3) into (1): \( (\alpha + \gamma) = n(\gamma - \beta) \)
\( \implies \alpha + \gamma = n\gamma - n\beta \)
\( \implies \alpha + n\beta = (n - 1)\gamma \) ... (Equation 4)
For paraxial rays, if \( h \) is the height of point M from the principal axis: \( \alpha \approx \tan \alpha = \frac{h}{u} \) \( \beta \approx \tan \beta = \frac{h}{v} \) \( \gamma \approx \tan \gamma = \frac{h}{R} \)
Substitute these small angle approximations into Equation (4): \( \frac{h}{u} + n\frac{h}{v} = (n - 1)\frac{h}{R} \)
Divide by \( h \) (assuming \( h \neq 0 \)): \( \frac{1}{u} + \frac{n}{v} = \frac{n - 1}{R} \)
This formula is correct when light travels from a rarer medium (like air, \( n_1=1 \)) to a denser medium (like glass, \( n_2=n \)). Using general refractive indices \( n_1 \) and \( n_2 \): \( \frac{n_1}{u} + \frac{n_2}{v} = \frac{n_2 - n_1}{R} \)
This is the formula for refraction at a convex spherical surface.
In simple words: When light goes from one medium to another through a curved surface, it bends. This formula, \( \frac{n_1}{u} + \frac{n_2}{v} = \frac{n_2 - n_1}{R} \), helps us find where the image will form. It uses the refractive index of both materials, the object distance, the image distance, and the curve of the surface.

🎯 Exam Tip: Carefully apply Snell's Law and small angle approximations. Remember that \( u \) is usually negative for real objects, \( v \) can be positive or negative, and \( R \) is positive for a convex surface from the perspective of incident light.

 

Question 5. Draw a diagram to illustrate the action of a compound microscope for the near point of the eye.
Answer: A compound microscope uses two lenses to magnify tiny objects, making them appear much larger than with a simple microscope. It consists of an objective lens (closer to the object, short focal length and small aperture) and an eyepiece (closer to the eye, larger focal length and aperture).
Figure 11.65 (not shown) illustrates the ray diagram for a compound microscope when the final image is formed at the least distance of distinct vision (near point).
Construction:The objective lens creates a real, inverted, and magnified intermediate image. This image then acts as the object for the eyepiece. The eyepiece, acting like a simple magnifying glass, further magnifies this intermediate image, forming a final image that is virtual, inverted, and greatly magnified.
Magnifying Power for Final Image at Least Distance of Distinct Vision (\( D \)):The magnifying power (\( M \)) is the ratio of the angle subtended by the final image at the eye (\( \beta \)) to the angle subtended by the object at the unaided eye (\( \alpha \)) when both are at the least distance of distinct vision.
\( M = \frac{\beta}{\alpha} \)
For small angles, \( M \approx \frac{\tan \beta}{\tan \alpha} \)
In the diagram (Fig 11.65, not shown), let \( A'B' \) be the intermediate image formed by the objective lens, and \( A''B'' \) be the final image formed by the eyepiece.
\( M = \frac{A'B'/u_e}{A'B'/D} = \frac{D}{u_e} \) (where \( u_e \) is object distance for eyepiece, and \( D \) is the least distance of distinct vision).
From the lens formula for the eyepiece: \( \frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e} \)
When the final image is formed at the least distance of distinct vision, \( v_e = -D \).
So, \( \frac{1}{-D} - \frac{1}{u_e} = \frac{1}{f_e} \)
\( \implies -\frac{1}{u_e} = \frac{1}{f_e} + \frac{1}{D} \)
\( \implies \frac{1}{u_e} = -(\frac{1}{f_e} + \frac{1}{D}) \)
Substitute \( \frac{1}{u_e} \) into the magnification formula \( M = \frac{D}{u_e} \): \( M = D \times (-(\frac{1}{f_e} + \frac{1}{D})) = -( \frac{D}{f_e} + \frac{D}{D} ) \)
\( \implies M = - (1 + \frac{D}{f_e}) \)
Also, for the objective lens, if \( v_o \) is the image distance and \( u_o \) is the object distance, the magnification of the objective is \( m_o = -\frac{v_o}{u_o} \). The total magnification is \( M = m_o \times m_e = (-\frac{v_o}{u_o}) (1 + \frac{D}{f_e}) \).
For the final image formed at infinity (normal adjustment), \( v_e = -\infty \), so \( \frac{1}{u_e} = -\frac{1}{f_e} \). Then \( M = - \frac{D}{f_e} \).
In this case, total magnification \( M = (-\frac{v_o}{u_o}) (\frac{D}{f_e}) \).
In simple words: A compound microscope helps see tiny things bigger. It uses two lenses. The first lens makes a medium-sized image, and the second lens magnifies that image even more for your eye. When you look at an object through it, the final image is very big, upside down, and seems to be at a comfortable distance for your eyes.

🎯 Exam Tip: Remember the two main cases for final image formation: at the least distance of distinct vision and at infinity. The formulas for magnifying power differ slightly. Always sketch a ray diagram to ensure you understand the image formation steps.

 

RBSE Class 12 Physics Chapter 11 Numerical Questions

 

Question 6. Draw a diagram to show refraction by prism for a monochromatic light source. Derive relation between angle of prism and minimum deviation angle in terms of refractive index of glass.
Answer: When a monochromatic (single color) light ray passes through a prism, it undergoes refraction (bending) two times. The diagram (Figure 11.47, not shown) illustrates a prism ABC with an incident ray PQ, refracted ray QR, and emergent ray RS.
Here, \( i_1 \) is the angle of incidence, \( r_1 \) and \( r_2 \) are angles of refraction inside the prism, \( i_2 \) (or \( e \)) is the angle of emergence, and \( \delta \) is the angle of deviation. \( A \) is the angle of the prism.
Relation between Angle of Prism and Minimum Deviation:Consider the quadrilateral AQNR (where Q and R are points on the prism faces where refraction occurs, and N is the intersection of normals). The sum of angles in a quadrilateral is 360°. \( \angle A + \angle N + \angle AQN + \angle ARN = 360^\circ \) Since normals are perpendicular to the faces: \( \angle AQN = 90^\circ \) and \( \angle ARN = 90^\circ \).
So, \( \angle A + \angle N + 90^\circ + 90^\circ = 360^\circ \)
\( \implies \angle A + \angle N = 180^\circ \) ... (Equation 1)
Now, consider the triangle QNR. The sum of angles in a triangle is 180°. \( \angle QNR + \angle NQR + \angle NRQ = 180^\circ \) \( \implies \angle N + r_1 + r_2 = 180^\circ \) ... (Equation 2)
Comparing Equation (1) and Equation (2): \( \angle A = r_1 + r_2 \) ... (Equation 3)
The angle of deviation \( \delta \) is the angle between the incident ray produced forward and the emergent ray produced backward. From the exterior angle property for the triangle formed by the incident, refracted, and emergent rays (let's call the intersection point X), \( \delta = (i_1 - r_1) + (i_2 - r_2) \)
\( \implies \delta = (i_1 + i_2) - (r_1 + r_2) \)
Substitute Equation (3) into this: \( \delta = i_1 + i_2 - A \) ... (Equation 4)
At Minimum Deviation (\( \delta_m \)):When the prism is in the position of minimum deviation, the angle of incidence equals the angle of emergence: \( i_1 = i_2 = i \). Also, the angles of refraction inside the prism become equal: \( r_1 = r_2 = r \).
Using Equation (3) for minimum deviation: \( A = r + r = 2r \implies r = \frac{A}{2} \) ... (Equation 5)
Using Equation (4) for minimum deviation: \( \delta_m = i + i - A = 2i - A \)
\( \implies 2i = A + \delta_m \implies i = \frac{A + \delta_m}{2} \) ... (Equation 6)
Now, applying Snell's Law at the first refracting surface (air to prism material): Refractive index \( n = \frac{\sin i_1}{\sin r_1} \)
At minimum deviation: \( n = \frac{\sin i}{\sin r} \)
Substitute \( i \) and \( r \) from Equations (6) and (5): \( n = \frac{\sin (\frac{A + \delta_m}{2})}{\sin (\frac{A}{2})} \)
This is the relation between the refractive index (\( n \)), the angle of prism (\( A \)), and the angle of minimum deviation (\( \delta_m \)) for a prism.
In simple words: When a single color of light goes through a prism, it bends twice. The amount it bends is called deviation. At its least bend (minimum deviation), the angle where light enters equals the angle where it leaves. Using simple geometry and Snell's law, we can find a formula that connects the prism's angle, the minimum bending angle, and how much the prism material bends light.

🎯 Exam Tip: The derivation for the prism formula is standard. Remember the two key geometric relations: \( A = r_1 + r_2 \) and \( \delta = i_1 + i_2 - A \). The conditions for minimum deviation (\( i_1 = i_2 \) and \( r_1 = r_2 \)) simplify these equations for the final result.

 

Question 6. Draw a diagram to show refraction by prism for a monochromatic light source. Derive relation between angle of prism and minimum deviation angle in terms of refractive index of glass.
Answer:
Refraction in Prism
When a light-ray passes through a prism, two main actions can happen: deviation and dispersion.

1. Deviation
When a monochromatic ray of light hits the refracting surface of the prism, it bends. The diagram below shows how light travels through a prism.

Here, PQ is the incident ray, QR is the refracted ray, and RS is the emergent ray. \(i_1\) is the angle of incidence, \(r_1\) and \(r_2\) are angles of refraction, \(i_2\) or \(e\) is the emergent angle, and \(\delta\) is the angle of deviation.

**Relation between angle of incidence, angle of deviation and angle of prism**
In \(\Delta AO'QR\), the sum of all three interior angles is 180°.
In the quadrilateral AQO'R, the sum of all interior angles is 360°.
\(\angle AQO' = 90^\circ\) and \(\angle ARO' = 90^\circ\) (because normals are perpendicular to the surfaces).
So, \(\angle QAR + \angle AQO' + \angle QO'R + \angle ARO' = 360^\circ\)
\(\angle A + 90^\circ + \angle QO'R + 90^\circ = 360^\circ\)
\(\angle A + \angle QO'R = 180^\circ\) ... (Equation 1)

Also, in \(\Delta QO'R\), the sum of angles is 180°:
\(r_1 + r_2 + \angle QO'R = 180^\circ\) ... (Equation 2)

Comparing (1) and (2):
\(A = r_1 + r_2\) ... (Equation 3)

The exterior angle of \(\Delta QTR\) (formed by extending PQ and RS) is \(\delta\).
So, \(\delta = (i_1 - r_1) + (i_2 - r_2)\)
\(\delta = i_1 + i_2 - (r_1 + r_2)\)
Substitute \(A = r_1 + r_2\) from (3):
\(\delta = i_1 + i_2 - A\) ... (Equation 4)

**Angle of Minimum Deviation (\(\delta_m\))**
When the deviation angle is minimum, the ray inside the prism is parallel to its base. In this condition:
\(i_1 = i_2 = i\) and \(r_1 = r_2 = r\)
From Equation (3): \(A = r + r = 2r \implies r = \frac{A}{2}\)
From Equation (4): \(\delta_m = i + i - A \implies \delta_m = 2i - A \implies i = \frac{A + \delta_m}{2}\)

**Formula for Refractive Index of the Prism**
According to Snell's Law, the refractive index (n) of the prism material is:
\(n = \frac{\sin i}{\sin r}\)
Substitute the values of \(i\) and \(r\):
\(n = \frac{\sin \left(\frac{A + \delta_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}\)
This formula is known as the **Prism Formula**.
In simple words: When light goes through a prism, it bends. The angle it bends by is called deviation. There's a special point where the bending is least (minimum deviation). Using the prism's angle and this minimum bending angle, we can find out how much the prism material bends light (its refractive index) using a specific formula.

🎯 Exam Tip: Remember the conditions for minimum deviation, especially \(i_1 = i_2\) and \(r_1 = r_2\), as they are key to deriving the prism formula correctly. Also, be careful with the angle definitions in the diagram.

 

Question 7. Assuming lens to be two spherical surface, derive relation between u, v and f.
Answer:
**Derivation of Lens Maker's Formula**
We consider a thin lens with two spherical surfaces. Let the medium on both sides of the lens be the same (e.g., air). The radii of curvature for the two surfaces are \(R_1\) and \(R_2\), respectively. A point object O is placed on the principal axis.

The image I' is formed by refraction through the first surface (\(AP_1B\)). This image I' then acts as a virtual object for the second surface (\(AP_2B\)), which forms the final image at I.

**For refraction through the first surface (\(AP_1B\))**
Using the refraction formula for a spherical surface: \(\frac{n_2}{v'} - \frac{n_1}{u} = \frac{n_2 - n_1}{R_1}\)
Here, \(n_1 = 1\) (for air) and \(n_2 = n\) (for lens material).
So, \(\frac{n}{v'} - \frac{1}{u} = \frac{n - 1}{R_1}\) ... (Equation 1)

**For refraction through the second surface (\(AP_2B\))**
The image I' from the first surface acts as a virtual object for the second surface. Its distance from the second surface is \(v' - t\), where t is the thickness of the lens. Since the lens is thin, \(t\) is negligible, so the object distance for the second surface is \(v'\).
For the second surface, light goes from lens material (refractive index \(n\)) back into air (refractive index 1). The radius of curvature is \(R_2\).
Using the refraction formula: \(\frac{n_1}{v} - \frac{n_2}{u'} = \frac{n_1 - n_2}{R_2}\)
Here, \(n_1 = 1\) (for air), \(n_2 = n\) (for lens material), \(u' = v'\) (virtual object for second surface), and \(R_2\) is usually taken as negative if the surface curves the other way.
So, \(\frac{1}{v} - \frac{n}{v'} = \frac{1 - n}{R_2}\)
\(\frac{1}{v} - \frac{n}{v'} = -\frac{n - 1}{R_2}\) ... (Equation 2)

**Adding Equation (1) and Equation (2):**
\(\left(\frac{n}{v'} - \frac{1}{u}\right) + \left(\frac{1}{v} - \frac{n}{v'}\right) = \frac{n - 1}{R_1} - \frac{n - 1}{R_2}\)
The terms \(\frac{n}{v'}\) and \(-\frac{n}{v'}\) cancel out:
\(\frac{1}{v} - \frac{1}{u} = (n - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)\)

**When the object is at infinity (\(u = \infty\)), the image is formed at the focal point (\(v = f\)).**
Substitute these into the equation:
\(\frac{1}{f} - \frac{1}{\infty} = (n - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)\)
\(\frac{1}{f} = (n - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)\)
This is the **Lens Maker's Formula**.

**Lens Equation (relation between u, v, and f):**
Comparing the derived formula with the previous step, we get:
\(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)
This is the **Lens Equation** or **Thin Lens Formula**.
In simple words: The Lens Maker's Formula helps us calculate the focal length of a lens based on its material (how much it bends light) and the curvature of its two surfaces. The Lens Equation then connects this focal length to where an object is placed and where its image will appear.

🎯 Exam Tip: Clearly state assumptions like "thin lens" and "same medium on both sides." Pay attention to sign conventions for \(R_1\) and \(R_2\) based on the lens type (convex/concave) and coordinate system chosen.

 

Question 8. Write the types of telescope. For a refractive telescope, derive relation for magnifying power. Also discuss about its construction and working.
Answer:
**Types of Telescopes**
Telescopes are optical instruments used to see distant objects clearly, which are otherwise not visible to the naked eye. Astronomers use them to observe celestial bodies like stars and planets. Based on their construction, telescopes are mainly of two types:
1. Refracting Telescope
2. Reflecting Telescope

**Refracting Type Telescope (Astronomical Telescope)**
**Construction:** A refracting telescope consists of two convex lenses mounted coaxially in a tube. These lenses are:
- **Objective Lens (\(L_o\)):** This is the lens towards the object. It has a large aperture and a long focal length (\(f_o\)).
- **Eyepiece Lens (\(L_e\)):** This is the lens towards the eye. It has a small aperture and a short focal length (\(f_e\)).
The distance between these lenses can be adjusted using a rack and pinion mechanism.

**Working and Image Formation (Normal Adjustment - Image at Infinity):**
When the telescope is in normal adjustment, the final image is formed at infinity. This is achieved when the intermediate image (\(A'B'\)) formed by the objective lens lies exactly at the focal point of the eyepiece lens (\(F_e\)).
The ray diagram for normal adjustment is shown below:

Parallel rays from a distant object (like a star) fall on the objective lens. It forms a real, inverted, and diminished image (\(A'B'\)) at its focal plane (\(F_o\)). This image is then observed by the eyepiece. In normal adjustment, the eyepiece is positioned so that \(A'B'\) lies exactly at its principal focus (\(F_e\)). The rays emerging from the eyepiece become parallel to each other, forming the final image at infinity. This relaxed viewing condition is preferred for long observations.

**Magnifying Power (M)**
The magnifying power of an astronomical telescope is defined as the ratio of the angle subtended by the final image at the eye (\(\beta\)) to the angle subtended by the object at the unaided eye (\(\alpha\)).
\(M = \frac{\beta}{\alpha}\)
For small angles, \(\beta \approx \tan \beta\) and \(\alpha \approx \tan \alpha\).
From the diagram:
\(\tan \alpha = \frac{A'B'}{A'O}\) (where O is the optical center of the objective lens)
\(\tan \beta = \frac{A'B'}{A'F_e}\) (where \(F_e\) is the focal point of the eyepiece)

Since \(A'O\) is the focal length of the objective lens (\(f_o\)), and \(A'F_e\) is the focal length of the eyepiece (\(f_e\)) (for normal adjustment), we have:
\(\tan \alpha = \frac{A'B'}{f_o}\)
\(\tan \beta = \frac{A'B'}{f_e}\)

Therefore, \(M = \frac{\tan \beta}{\tan \alpha} = \frac{A'B'/f_e}{A'B'/f_o} = \frac{f_o}{f_e}\)
In normal adjustment, the image is inverted, so a negative sign is often added to indicate this: \(M = -\frac{f_o}{f_e}\)

**Length of the Telescope (L):**
In normal adjustment, the distance between the objective lens and the eyepiece is \(L = f_o + f_e\).

**Magnifying Power when the Final Image Forms at the Least Distance of Distinct Vision (D):**
When the final image forms at the least distance of distinct vision (D = 25 cm), the eyepiece acts like a simple microscope. For the eyepiece, the image distance \(v_e = -D\) and the object distance is \(u_e\).
Using the lens formula for the eyepiece: \(\frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e}\)
\(\frac{1}{-D} - \frac{1}{u_e} = \frac{1}{f_e}\)
\(\frac{1}{u_e} = -\frac{1}{D} - \frac{1}{f_e} = -\left(\frac{1}{D} + \frac{1}{f_e}\right) = -\frac{f_e + D}{Df_e}\)
So, \(u_e = -\frac{Df_e}{D + f_e}\)

The magnifying power of the eyepiece in this case is \(M_e = \left(1 + \frac{D}{f_e}\right)\).
The magnifying power of the objective lens is \(M_o = \frac{v_o}{u_o}\). Since the object is at infinity, \(u_o = \infty\), but the image formed by the objective \(A'B'\) is at \(v_o = f_o\). The angular magnification is still primarily \(f_o/u_e\).
The total magnifying power is \(M = \frac{f_o}{u_e}\). Replacing \(u_e\),
\(M = \frac{f_o}{\left(\frac{Df_e}{D+f_e}\right)} = \frac{f_o}{f_e}\left(1 + \frac{f_e}{D}\right)\) (ignoring negative sign for magnitude)
The total magnifying power of the telescope for least distance of distinct vision is \(M = \frac{f_o}{f_e}\left(1 + \frac{f_e}{D}\right)\).
In simple words: Telescopes help us see far-away things. Refracting telescopes use two lenses: a big one (objective) to collect light from the distant object and a small one (eyepiece) to magnify the image for our eye. The magnifying power tells us how much bigger the distant object appears, and it depends on the focal lengths of these two lenses.

🎯 Exam Tip: Distinguish clearly between normal adjustment (image at infinity, relaxed eye) and image at least distance of distinct vision (strained eye). The ray diagrams and formulas for magnifying power and tube length differ for these two cases.

RBSE Class 12 Physics Chapter 11 Numerical Questions

 

Question 1. An object is placed at 36 cm in front of concave mirror of focal length 24 cm. Find out the distance of image formed.
Answer:
Given:
Focal length of concave mirror, \(f = -24\) cm (negative for concave mirror)
Object distance, \(u = -36\) cm (negative as object is in front of mirror)
Image distance, \(v = ?\)

Using the mirror formula: \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\)
Substituting the given values:
\(\frac{1}{-24} = \frac{1}{v} + \frac{1}{-36}\)
\(\frac{1}{v} = \frac{1}{-24} - \frac{1}{-36}\)
\(\frac{1}{v} = \frac{1}{36} - \frac{1}{24}\)
To find a common denominator, which is 72:
\(\frac{1}{v} = \frac{2}{72} - \frac{3}{72}\)
\(\frac{1}{v} = \frac{2 - 3}{72}\)
\(\frac{1}{v} = \frac{-1}{72}\)
\(v = -72\) cm

The negative sign indicates that the image is formed on the same side as the object (in front of the mirror), meaning it is a real image. The image is formed at 72 cm from the mirror.
In simple words: When an object is placed in front of a concave mirror, we can use a formula to find where the image will appear. For this mirror and object, the image forms 72 cm in front of the mirror, which means it is a real image.

🎯 Exam Tip: Always use the correct sign conventions for focal length, object distance, and image distance when applying the mirror formula to avoid errors. Negative signs typically denote real objects/images or distances measured against the direction of incident light.

 

Question 2. The relative refractive index of any medium in vacuum is 1.33. The velocity of light in vacuum is \(c = 3 \times 10^8\) m/s. Then find out velocity of light in medium.
Answer:
Given:
Relative refractive index of the medium, \(n_m = 1.33\) (or \(n_{air} = 1.33\))
Velocity of light in vacuum, \(c = 3 \times 10^8\) m/s
Velocity of light in the medium, \(v_m = ?\)

The refractive index of a medium is defined as the ratio of the speed of light in vacuum to the speed of light in the medium:
\(n_m = \frac{c}{v_m}\)
Rearrange the formula to find \(v_m\):
\(v_m = \frac{c}{n_m}\)
Substitute the given values:
\(v_m = \frac{3 \times 10^8 \text{ m/s}}{1.33}\)
\(v_m \approx 2.25 \times 10^8 \text{ m/s}\)
Therefore, the velocity of light in the given medium is approximately \(2.25 \times 10^8\) m/s.
In simple words: Light travels slower in materials than in empty space. The refractive index tells us how much slower. Knowing the speed of light in empty space and the material's refractive index, we can calculate how fast light moves through that material.

🎯 Exam Tip: Remember the basic definition of refractive index as \(n = c/v\). Ensure units are consistent (m/s for speed) and perform division carefully, especially with large numbers in scientific notation.

 

Question 3. The radii of curvatures for a convex lens of a focal length 20 cm are 18 cm and 24 cm. What is the refractive index of the glass?
Answer:
Given:
Focal length of the convex lens, \(f = +20\) cm (positive for convex lens)
Radius of curvature of the first surface, \(R_1 = +18\) cm (convex surface, incident light from left)
Radius of curvature of the second surface, \(R_2 = -24\) cm (convex surface facing right has its center on the left of vertex, so \(R_2\) is negative)
Refractive index of the glass, \(n = ?\)

Using the Lens Maker's Formula:
\(\frac{1}{f} = (n - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)\)
Substitute the given values:
\(\frac{1}{20} = (n - 1) \left(\frac{1}{18} - \frac{1}{-24}\right)\)
\(\frac{1}{20} = (n - 1) \left(\frac{1}{18} + \frac{1}{24}\right)\)
Find a common denominator for 18 and 24, which is 72:
\(\frac{1}{20} = (n - 1) \left(\frac{4}{72} + \frac{3}{72}\right)\)
\(\frac{1}{20} = (n - 1) \left(\frac{4 + 3}{72}\right)\)
\(\frac{1}{20} = (n - 1) \left(\frac{7}{72}\right)\)
Now, solve for \((n - 1)\):
\(n - 1 = \frac{1}{20} \times \frac{72}{7}\)
\(n - 1 = \frac{72}{140}\)
\(n - 1 = \frac{18}{35}\)
\(n - 1 \approx 0.514\)
\(n = 1 + 0.514\)
\(n \approx 1.514\)
The refractive index of the glass is approximately 1.514.
In simple words: We can figure out how much a lens material bends light (its refractive index) if we know its focal length and the curved shapes of its surfaces. Using a special formula and the given numbers, the glass in this lens has a refractive index of about 1.514.

🎯 Exam Tip: Correctly applying sign conventions for \(R_1\) and \(R_2\) is crucial. For a biconvex lens, the first surface \(R_1\) is usually positive, and the second surface \(R_2\) is negative, assuming light comes from the left.

 

Question 4. A ray of light is incident on a glass plate at 50°. If angle of refraction is 30°, then find out refractive index of glass.
Answer:
Given:
Angle of incidence, \(i = 50^\circ\)
Angle of refraction, \(r = 30^\circ\)
Refractive index of glass, \(n = ?\)

Using Snell's Law:
\(n = \frac{\sin i}{\sin r}\)
Substitute the given values:
\(n = \frac{\sin 50^\circ}{\sin 30^\circ}\)
We know \(\sin 50^\circ \approx 0.766\) and \(\sin 30^\circ = 0.5\)
\(n = \frac{0.766}{0.5}\)
\(n = 1.532\)
The refractive index of the glass is 1.532.
In simple words: When light passes from one material to another, it bends. We can measure how much it bends (angles of incidence and refraction) and then use a rule called Snell's Law to calculate the refractive index of the material, which tells us its light-bending ability.

🎯 Exam Tip: Remember Snell's Law in the form \(n = \sin i / \sin r\). Ensure your calculator is in degree mode for trigonometric functions. Accurate sine values are important for the final answer.

 

Question 5. An object is at 0.06 m from a convex lens of focal length 0.10 m. Find out the position of the image.
Answer:
Given:
Object distance, \(u = -0.06\) m (negative as object is in front of the lens)
Focal length of convex lens, \(f = +0.10\) m (positive for convex lens)
Image distance, \(v = ?\)

Using the lens formula:
\(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\)
Substitute the given values:
\(\frac{1}{0.10} = \frac{1}{v} - \frac{1}{-0.06}\)
\(\frac{1}{0.10} = \frac{1}{v} + \frac{1}{0.06}\)
\(\frac{1}{v} = \frac{1}{0.10} - \frac{1}{0.06}\)
Convert decimals to fractions or work with integers by multiplying by 100:
\(\frac{1}{v} = \frac{10}{1} - \frac{100}{6}\)
\(\frac{1}{v} = 10 - \frac{50}{3}\)
\(\frac{1}{v} = \frac{30}{3} - \frac{50}{3}\)
\(\frac{1}{v} = \frac{30 - 50}{3}\)
\(\frac{1}{v} = \frac{-20}{3}\)
\(v = -\frac{3}{20}\) m
\(v = -0.15\) m (or -15 cm)
The negative sign indicates that the image is formed on the same side as the object (virtual image). The image is formed 0.15 m from the lens.
In simple words: With a convex lens, an object placed at a certain distance will form an image. Using the lens formula, we find that the image is formed 0.15 meters away from the lens, on the same side as the object, meaning it is a virtual image.

🎯 Exam Tip: Always use the correct sign conventions for object distance (u) and focal length (f) based on the lens type. For a convex lens, f is positive; for an object in front, u is negative. Remember the lens formula: \(1/f = 1/v - 1/u\).

 

Question 7. Two lenses of power +5 D and -7 D are in contact with each other. Find out the power of the combination of the lenses. Combined lens will be convex or concave?
Answer:
Given:
Power of the first lens, \(P_1 = +5\) D
Power of the second lens, \(P_2 = -7\) D
Combined power, \(P = ?\)

For lenses in contact, the total power of the combination is the algebraic sum of their individual powers:
\(P = P_1 + P_2\)
Substitute the given values:
\(P = +5 \text{ D} + (-7 \text{ D})\)
\(P = 5 - 7\)
\(P = -2\) D

Since the combined power \(P\) is negative (\(-2\) D), the combination behaves like a concave lens. A concave lens has a diverging effect on light.
In simple words: When two lenses are put together, their powers add up. If one lens has a positive power and the other a negative power, their combined power might be negative, which means the new setup acts like a concave lens that spreads light out.

🎯 Exam Tip: Remember that power for lenses in contact is simply additive. A positive total power indicates a convex (converging) combination, while a negative total power indicates a concave (diverging) combination.

 

Question 8. The focal length of objective lens and eyepiece for a compound microscope are 0.95 cm and 5 cm, and they are placed at 20 cm from each other. Final image is formed at 25 cm in eyepiece. Find out the magnifying power of the microscope.
Answer:
Given:
Focal length of objective lens, \(f_o = 0.95\) cm
Focal length of eyepiece, \(f_e = 5\) cm
Distance between objective and eyepiece, \(L = |v_o| + |u_e| = 20\) cm
Final image formed at \(v_e = -25\) cm (least distance of distinct vision, negative by convention)

**Step 1: Find the object distance for the eyepiece (\(u_e\))**
Using the lens formula for the eyepiece: \(\frac{1}{f_e} = \frac{1}{v_e} - \frac{1}{u_e}\)
\(\frac{1}{5} = \frac{1}{-25} - \frac{1}{u_e}\)
\(\frac{1}{u_e} = \frac{1}{-25} - \frac{1}{5}\)
\(\frac{1}{u_e} = \frac{-1}{25} - \frac{5}{25}\)
\(\frac{1}{u_e} = \frac{-1 - 5}{25}\)
\(\frac{1}{u_e} = \frac{-6}{25}\)
\(u_e = -\frac{25}{6}\) cm
The magnitude \(|u_e| = \frac{25}{6}\) cm.

**Step 2: Find the image distance for the objective lens (\(v_o\))**
We know that \(L = |v_o| + |u_e|\)
\(20 = |v_o| + \frac{25}{6}\)
\(|v_o| = 20 - \frac{25}{6}\)
\(|v_o| = \frac{120 - 25}{6}\)
\(|v_o| = \frac{95}{6}\) cm
For a compound microscope, the image from the objective is real and inverted, so \(v_o\) is positive: \(v_o = \frac{95}{6}\) cm.

**Step 3: Find the object distance for the objective lens (\(u_o\))**
Using the lens formula for the objective lens: \(\frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o}\)
\(\frac{1}{0.95} = \frac{1}{\frac{95}{6}} - \frac{1}{u_o}\)
\(\frac{1}{0.95} = \frac{6}{95} - \frac{1}{u_o}\)
\(\frac{1}{u_o} = \frac{6}{95} - \frac{1}{0.95}\)
\(\frac{1}{u_o} = \frac{6}{95} - \frac{100}{95}\)
\(\frac{1}{u_o} = \frac{6 - 100}{95}\)
\(\frac{1}{u_o} = \frac{-94}{95}\)
\(u_o = -\frac{95}{94}\) cm

**Step 4: Calculate the total magnifying power (M)**
The magnifying power for a compound microscope when the final image is formed at the least distance of distinct vision is given by:
\(M = M_o \times M_e = \left(\frac{v_o}{|u_o|}\right) \times \left(1 + \frac{D}{f_e}\right)\)
\(M = \left(\frac{\frac{95}{6}}{\frac{95}{94}}\right) \times \left(1 + \frac{25}{5}\right)\)
\(M = \left(\frac{95}{6} \times \frac{94}{95}\right) \times (1 + 5)\)
\(M = \frac{94}{6} \times 6\)
\(M = 94\)
The total magnifying power of the microscope is 94.
In simple words: For a compound microscope, we can find its total magnifying power by calculating how much each lens (objective and eyepiece) magnifies the image. We use the focal lengths of the lenses and the distances to the object and images to combine these individual magnifications into one final number, which is 94 in this case.

🎯 Exam Tip: In compound microscope problems, carefully track the sign conventions and ensure the image formed by the objective acts as the object for the eyepiece. Remember the formula for total magnification as the product of individual lens magnifications.

 

Question 9. The power of a thin convex lens (\(n_g = 1.5\)) is +5.0D, when this lens is dipped in a liquid of refractive index \(n_l\), it behaves like a concave lens whose focal length is 100 cm, then what will be the value of \(n_l\)?
Answer:
Given:
Refractive index of glass (lens material), \(n_g = 1.5\)
Power of the lens in air, \(P_a = +5.0\) D
Focal length of the lens in liquid, \(f_l = -100\) cm (negative because it behaves like a concave lens)
Refractive index of the liquid, \(n_l = ?\)

**Step 1: Find the focal length of the lens in air (\(f_a\))**
We know that power \(P = \frac{1}{f \text{ (in meters)}}\). So, \(f_a = \frac{1}{P_a}\)
\(f_a = \frac{1}{+5.0 \text{ D}} = +0.20\) m = \(+20\) cm

**Step 2: Use the Lens Maker's Formula for the lens in air**
\(\frac{1}{f_a} = (n_g - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)\)
\(\frac{1}{20} = (1.5 - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)\)
\(\frac{1}{20} = (0.5) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)\)
So, \(\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \frac{1}{20 \times 0.5} = \frac{1}{10}\) cm\(^{-1}\) ... (Equation 1)

**Step 3: Use the Lens Maker's Formula for the lens in liquid**
When the lens is dipped in a liquid, the refractive index of the lens material relative to the liquid is \(\frac{n_g}{n_l}\).
The formula becomes: \(\frac{1}{f_l} = \left(\frac{n_g}{n_l} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)\)
We are given \(f_l = -100\) cm.
\(\frac{1}{-100} = \left(\frac{1.5}{n_l} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)\)

**Step 4: Substitute Equation (1) into the liquid formula**
\(\frac{1}{-100} = \left(\frac{1.5}{n_l} - 1\right) \left(\frac{1}{10}\right)\)
Multiply both sides by 10:
\(\frac{10}{-100} = \frac{1.5}{n_l} - 1\)
\(-\frac{1}{10} = \frac{1.5}{n_l} - 1\)
\(1 - \frac{1}{10} = \frac{1.5}{n_l}\)
\(\frac{10 - 1}{10} = \frac{1.5}{n_l}\)
\(\frac{9}{10} = \frac{1.5}{n_l}\)
Now, solve for \(n_l\):
\(n_l = \frac{1.5 \times 10}{9}\)
\(n_l = \frac{15}{9}\)
\(n_l = \frac{5}{3}\)
\(n_l \approx 1.67\)
The refractive index of the liquid is approximately 1.67.
In simple words: A lens usually bends light in one way, but if you put it into a liquid with a different light-bending ability, it can start bending light the opposite way. By knowing the lens's power in air and its new focal length in the liquid, we can calculate the liquid's light-bending ability (refractive index).

🎯 Exam Tip: This problem involves two scenarios (air and liquid) and requires careful application of the Lens Maker's Formula. The term \(\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\) remains constant for the lens. Remember to convert power to focal length and apply correct sign conventions for focal lengths in both mediums.

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