CBSE Class 12 Physics Optics Assignment Set 02

VERY SHORT ANSWER QUESTIONS (1 Mark)

Question 1. Write the value of angle of reflection for a ray of light falling normally on a mirror.
Answer: Zero
In simple words: When light strikes a mirror perpendicularly, the angle of incidence is \(0^\circ\), so the angle of reflection is also \(0^\circ\).

Exam Tip: Remember the laws of reflection: angle of incidence equals angle of reflection (\(i = r\)).

Question 2. How does the dispersive power of glass prism change when it is dipped in water?
Answer: Decreases.
In simple words: Dispersive power depends on the relative refractive index of the prism material with respect to the surrounding medium, which decreases when placed in water.

Exam Tip: Dispersive power \(\omega = \frac{\mu_v - \mu_r}{\mu - 1}\). When dipped in water, relative refractive index decreases, hence \(\omega\) decreases.

Question 3. Light falls from glass to air. Find the angle of incidence for which the angle of deviation is \(90^\circ\).
Answer: Angle of incidence is \(45^\circ\).
In simple words: Deviation \(\delta = r - i\). For total internal reflection or specific deviation cases, substitute \(\delta = 90^\circ\). Assuming simple refraction where ray grazes the surface (\(r = 90^\circ\)), deviation \(\delta = 90^\circ - i\), giving \(i = 45^\circ\) for standard glass refractive index \(\sqrt{2}\).

Exam Tip: Use the relation for deviation during refraction: \(\delta = r - i\).

Question 4. Name the phenomenon due to which one cannot see through fog.
Answer: Scattering.
In simple words: Water droplets in fog scatter light in all directions, preventing clear image formation.

Exam Tip: Mie scattering or general light scattering causes reduction in visibility through fog and clouds.

Question 5. What is the ratio of \(\sin i\) and \(\sin r\) in terms of velocities in the given figure?
Answer: \(\frac{\sin i}{\sin r} = \frac{v_1}{v_2} = \frac{\mu_2}{\mu_1}\)
In simple words: According to Snell's law, the ratio of sines of angles is equal to the ratio of speeds of light in the respective media.

Exam Tip: Always relate refractive index inversely with speed: \(\mu = \frac{c}{v}\).

Question 6. What is the shape of fringes in young's double slit experiment?
Answer: Hyperbolic.
In simple words: The intersection of spherical wave fronts with a flat screen produces hyperbolic interference fringes, though they appear nearly straight in the central region.

Exam Tip: Exact theoretical shape is hyperbola, but approximated as straight equidistant bands in numerical problems.

Question 7. A equiconcave lens of focal length \(15\text{ cm}\) is cut into two equal halves along dotted line as shown in figure. What will be new focal length of each half?
Answer: \(30\text{ cm}\)
In simple words: Cutting a lens perpendicular to its principal axis doubles the focal length of each resulting half-lens because optical power is halved.

Exam Tip: Power \(P \propto \frac{1}{f}\). When thickness is halved along the principal section, focal length becomes \(2f\).

Question 8. For the same angle of the incidence the angle of refraction in three media A, B and C are \(15^\circ\), \(25^\circ\) and \(35^\circ\) respectively. In which medium would the velocity of light be minimum?
Answer: Medium A.
In simple words: Smaller angle of refraction means higher refractive index, which corresponds to the lowest speed of light.

Exam Tip: Speed \(v \propto \sin r\) for a constant angle of incidence \(i\).

Question 9. What is the phase difference between two points on a cylindrical wave front?
Answer: Zero.
In simple words: All points lying on the same wavefront oscillate in exact same phase, so phase difference is zero.

Question 10. What is the 'power' of plane glass plate.
Answer: Zero.
In simple words: A plane glass plate has infinite focal length, and optical power is reciprocal of focal length, making it zero.

Exam Tip: \(P = \frac{1}{f} = \frac{1}{\infty} = 0\).

Question 11. Show with the help of diagram, why a beam of white light passing through a hollow prism does not give spectrum.
Answer: A hollow prism contains air inside. Refraction occurs only at the thin glass walls where emergent rays from the first refracting surface are parallel and opposite to those at the second surface, resulting in zero net dispersion and deviation.
In simple words: Dispersion requires a denser medium where different wavelengths travel at different speeds and bend by different angles. Air inside a hollow prism does not cause dispersion.

Exam Tip: Emphasize that dispersion is a property of material medium refractive index variation with wavelength.

Question 12. How does focal length of lens change when red light incident on it is replaced by violet light?
Answer: Decreases.
In simple words: Violet light has a higher refractive index than red light in glass, causing it to bend more and thus have a shorter focal length.

Exam Tip: Use Lens Maker's Formula: \(\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\). Since \(\mu_v > \mu_r\), \(f_v < f_r\).

Question 13. A myopic person prefers to remove his spectacles while reading a book. Why?
Answer: A myopic person is short-sighted; their near point is very close. Without spectacles, they can comfortably focus on near objects within their unassisted near point.
In simple words: Myopia helps in seeing nearby objects clearly without any corrective diverging lenses.

Exam Tip: Myopia means distant objects are blurry, but close-up vision remains clear.

Question 14. Lower half of the concave mirror is painted black. What effect will this have on the image of an object placed in front of the mirror?
Answer: The intensity of the image will be reduced (in this case half) but no change in size of the image.
In simple words: Every part of the mirror forms a complete image. Blocking half the mirror only reduces the amount of light reflecting, making the image dimmer, not smaller.

Question 1. A near sighted person can clearly see objects up to a distance of \(1.5\text{ m}\). Calculate power of the lens necessary for the remedy of this defect. (\(P = -0.67\text{ D}\))
Answer: Given object distance \(u = -\infty\), image distance \(v = -1.5\text{ m}\).
Using lens formula: \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\)
\(\frac{1}{f} = \frac{1}{-1.5} - \frac{1}{-\infty} = -\frac{1}{1.5} = -\frac{2}{3}\text{ m}^{-1}\)
Power \(P = \frac{1}{f} = -\frac{2}{3} = -0.67\text{ D}\).
In simple words: A diverging lens of power \(-0.67\text{ D}\) is required to correct myopia for a \(1.5\text{ m}\) far point.

Exam Tip: Always use sign convention properly: far point corresponds to image distance \(v\).

Question 2. A person can adjust the power of his eye lens between \(50\text{ D}\) and \(60\text{ D}\). His far point is infinity. Find the distance between retina and eye lens.
Answer: For far point at infinity (\(u = -\infty\)), power is minimum \(P = 50\text{ D}\).
Focal length \(f = \frac{1}{P} = \frac{1}{50}\text{ m} = 0.02\text{ m} = 2\text{ cm}\).
Since image is formed on the retina for far objects, the distance between retina and lens equals this focal length, which is \(2\text{ cm}\).
In simple words: The distance between the eye lens and retina is fixed at \(2\text{ cm}\), corresponding to the minimum power of \(50\text{ D}\).

Question 3. Calculate the value of \(\theta\), for which light incident normally on face AB grazes along the face BC. Given \(\mu_{\text{glass}} = 3/2\) and \(\mu_{\text{water}} = 4/3\).
For total internal reflection at face BC, angle of incidence inside glass equals critical angle \(i_c\).
\(\sin i_c = \frac{\mu_{\text{water}}}{\mu_{\text{glass}}} = \frac{4/3}{3/2} = \frac{8}{9}\).
From geometry of the prism, \(\theta = \sin^{-1}\left(\frac{8}{9}\right)\).
Answer: \(\theta = \sin^{-1}(8/9)\)
In simple words: Apply Snell's law at the interface and use the critical angle condition for grazing emergence.

Exam Tip: Critical angle \(\sin i_c = \frac{\mu_{\text{rarer}}}{\mu_{ denser}}\).

Question 4. Name any two characteristics of light which do not change on polarisation.
Answer: Speed and frequency.
In simple words: When light gets polarized, its frequency and speed in a medium remain unchanged, whereas intensity and amplitude change.

Question 5. Complete the path of light with correct value of angle of emergence.
Answer: \(\theta = \sin^{-1}(3/4)\)
In simple words: Trace the ray using Snell's law at each boundary until it emerges into air.

Exam Tip: Reversibility of light and Snell's law are key to completing ray paths.

Question 6. Define diffraction. What should be the order of the size of the aperture to observe diffraction.
Answer: Diffraction is the phenomenon of bending of light around the corners of an obstacle or aperture into the region of geometrical shadow.
The size of the aperture or obstacle must be of the order of the wavelength of light (\(\lambda\)) to observe prominent diffraction.
In simple words: Light bends around small obstacles whose size matches the wavelength of the light wave.

Exam Tip: Order of aperture size \(\approx \lambda\) (hundreds of nanometers for visible light).

Question 7. Show that maximum intensity in interference pattern is four times the intensity due to each slit if amplitude of light emerging from slits is same.
Answer: Let amplitude of each wave be \(a\).
Maximum intensity \(I_{\max} \propto (a_1 + a_2)^2 = (a + a)^2 = (2a)^2 = 4a^2\).
Intensity due to each slit \(I_0 \propto a^2\).
Therefore, \(I_{\max} = 4I_0\).
In simple words: When amplitudes add up constructively, maximum intensity becomes four times the individual slit intensity.

Question 8. Two poles-one \(4\text{ m}\) high and the other \(4.5\text{ m}\) high are situated at distance \(40\text{ m}\) and \(50\text{ m}\) respectively from an eye. Which pole will appear taller?
Answer: The first pole (\(4\text{ m}\) high at \(40\text{ m}\)) will appear taller.
In simple words: Visual angle subtended by the first pole is \(\frac{4}{40} = 0.1\text{ rad}\), while for the second pole it is \(\frac{4.5}{50} = 0.09\text{ rad}\). Larger angle means it appears taller.

Exam Tip: Apparent height is determined by the angle subtended at the eye (\(\theta \approx \frac{\text{height}}{\text{distance}}\)).

Question 9. \(S_1\) and \(S_2\) are two sources of light separated by a distance \(d\). A detector can move along \(S_2P\) perpendicular to \(S_1S_2\). What should be the minimum and maximum path difference at the detector?
Answer: Minimum path difference is zero (when point P is at infinity).
Maximum path difference is \(d\).
In simple words: As the detector moves from infinity towards the sources, the path difference increases from zero up to the separation distance \(d\).

Question 10. If a jogger runs with constant speed towards a vehicle, how fast does the image of the jogger appear to move in the rear view mirror when (i) the vehicle is stationery (ii) the vehicle is moving with constant speed.
Answer: (i) The speed of the image appears to increase substantially as the jogger gets closer to the vehicle.
(ii) A similar non-linear increase in image speed is observed when the vehicle is in motion.
In simple words: Convex mirrors compress distance, so as objects get closer, their images move across the mirror much faster.

Exam Tip: Mirror magnification changes non-linearly with object distance.

Question 11. A person looking at a mesh of crossed wire is able to see the vertical wire more distinctly than the horizontal wire. Which defect he is suffering from? How can this defect be corrected?
Answer: Astigmatism. It can be corrected using cylindrical lenses.
In simple words: Astigmatism happens when the cornea is not spherical, causing different curvatures in horizontal and vertical planes. Cylindrical lenses correct this imbalance.

Question 12. Is optical density same as mass density? Give an example.
Answer: No. Optical density is the ratio of the speed of light in two media, whereas mass density is mass per unit volume of a substance.
For example, mass density of turpentine oil is less than that of water, but its optical density is higher.
In simple words: Optical density measures how slow light travels in a medium, which is independent of its physical mass density.

Exam Tip: Do not confuse optical refractive index with physical mass density (e.g., kerosene/turpentine float on water but have higher refractive index).

Question 13. When does (i) a plane mirror and (ii) a convex mirror produce real image of objects.
Answer: Plane and convex mirrors produce real images when the object is virtual—that is, when converging rays meeting behind the mirror are reflected to a point on a screen.
In simple words: Real images are formed by plane and convex mirrors only when incoming light rays are already converging towards a virtual object point.

Exam Tip: Real object always forms virtual image in plane and convex mirrors.

Question 14. A virtual image cannot be caught on a screen. Then how do we see it?
Answer: The image is virtual when reflected or refracted rays diverge; these are converged onto the retina by the convex lens of the eye, as the virtual image serves as the object.
In simple words: Our eye lens converges the diverging rays coming from a virtual image to form a real image on our retina.

Exam Tip: The eye lens treats virtual images as real objects placed at the virtual image position.

Question 15. Draw a diagram to show the advance sunrise and delayed sunset due to atmospheric refraction. (NCERT Pg 318)
Answer: Refer to NCERT textbook figure on atmospheric refraction where light rays bend downwards through atmospheric layers, making the sun visible \(2\) minutes before actual sunrise and after sunset.
In simple words: Earth's atmosphere bends light rays from the sun, making it appear above the horizon even when it is geometrically below it.

Exam Tip: Total time difference due to advance sunrise and delayed sunset is approximately \(4\) minutes.

Question 16. Define critical angle for total internal reflection. Obtain an expression for refractive index of the medium in terms of critical angle.
Answer: Critical angle is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is \(90^\circ\).
By Snell's law: \(\mu_1 \sin i_c = \mu_2 \sin 90^\circ\)
If rarer medium is air (\(\mu_2 = 1\)), then \(\mu = \frac{1}{\sin i_c}\).
In simple words: The critical angle is the specific incidence angle where the refracted ray travels right along the boundary surface.

Question 17. The image of a small bulb fixed on the wall of a room is to be obtained on the opposite wall 's' m away by means of a large convex lens. What is the maximum possible focal length of the lens required?
Answer: For a fixed distance \(s\) between object and screen, for the lens equation to give real solution for \(u = v = 2f\), \(f\) should not be greater than \(4f = s\).
Therefore, maximum focal length \(f = \frac{s}{4}\).
In simple words: To focus an image on a wall using a convex lens, the distance between object and screen must be at least four times the focal length.

Exam Tip: Minimum distance between a real object and its real image in a convex lens is \(4f\).

Question 18. The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does magnifying glass produce angular magnification?
Answer: The absolute image size is bigger than the object size, the magnifier helps in bringing the object closer to the eye and hence it has a larger angular size than the same object at \(25\text{ cm}\), thus angular magnification is achieved.
In simple words: A magnifier allows us to bring an object much closer than the normal near point (\(25\text{ cm}\)), increasing the angle it subtends at the eye.

Question 19. Obtain relation between focal length and radius of curvature of (i) concave mirror (ii) convex mirror using proper ray diagram.
Answer: Using geometry of reflection for paraxial rays, the incident ray parallel to principal axis passes through (or appears to diverge from) the principal focus after reflection. From triangle geometry, \(R = 2f\) or \(f = R/2\).
In simple words: The focal length of a spherical mirror is always half of its radius of curvature for paraxial rays.

Exam Tip: Standard formula: \(f = \frac{R}{2}\) holds true for both concave and convex mirrors under paraxial approximation.

Question 20. Two independent light sources cannot act as coherent sources. Why?
Answer: Light is emitted by individual atoms undergoing transitions lasting about \(10^{-8}\) seconds. Independent sources emit light waves with random and rapidly changing phase differences, making them incoherent.
In simple words: Coherent sources require a constant phase relationship, which independent atoms cannot maintain.

Question 21. How is a wave front different from a ray? Draw the geometrical shape of the wavefronts when: (i) light diverges from a point source, (ii) light emerges out of convex lens when a point source is placed at its focus.
Answer: A wavefront is a surface of constant phase, whereas a ray is a line drawn perpendicular to the wavefront indicating the direction of light propagation.
(i) Spherical wavefront.
(ii) Plane wavefront.
In simple words: Wavefronts are surfaces where all light waves are in step, while rays show the direction they travel. Point sources create spherical wavefronts, and focused lenses create plane wavefronts.

Question 22. What two main changes in diffraction pattern of single slit will you observe when the monochromatic source of light is replaced by a source of white light?
Answer: (i) Each diffraction order shows component colors of white light. Since fringe width \(\alpha \lambda\), red fringes are wider than violet fringes.
(ii) In higher-order spectra, dispersion increases, causing overlapping of different colors.
In simple words: White light produces a colored central white maximum bordered by colored diffraction fringes due to different wavelengths.

Exam Tip: Fringe width is directly proportional to wavelength (\(\beta \propto \lambda\)), so red spreads out more than violet.

Question 23. You are provided with four convex lenses of focal length \(1\text{ cm}\), \(3\text{ cm}\), \(10\text{ cm}\) and \(100\text{ cm}\). Which two would you prefer for a microscope and which two for a telescope?
Answer: For Microscope: Objective \(f_o = 1\text{ cm}\) and Eyepiece \(f_e = 3\text{ cm}\).
For Telescope: Objective \(f_o = 100\text{ cm}\) and Eyepiece \(f_e = 1\text{ cm}\).
In simple words: Microscopes need very small focal length lenses for high magnification, while telescopes need a large objective and small eyepiece.

Question 24. Give reasons for the following: (i) Sun looks reddish at sunset, (ii) clouds are generally white.
Answer: (i) At sunset, light travels a longer path through the atmosphere. Shorter wavelengths (blue/violet) scatter away completely, leaving longer red wavelengths to reach our eyes.
(ii) Clouds contain water droplets and dust particles whose size is much larger than the wavelength of light, causing Mie scattering where all wavelengths scatter equally, appearing white.
In simple words: Red light survives long atmospheric travel at sunset, while clouds scatter all colors equally to look white.

Exam Tip: Rayleigh scattering explains red sun (\(I \propto \frac{1}{\lambda^4}\)), while Mie scattering explains white clouds.

Question 25. Using Huygens Principle draw ray diagram for the following: (i) Refraction of a plane wave front incident on a rarer medium, (ii) Refraction of a plane wave front incident on a denser medium.
Answer: Refer to NCERT Fig. 10.5 and Fig. 10.4 illustrating Huygens construction for refraction at plane surfaces.
In simple words: Secondary wavelets construct refracted wavefronts demonstrating Snell's law based on wave speed differences.

Question 1. Using mirror formula show that virtual image produced by a convex mirror is always smaller in size and is located between the focus and the pole.
Answer: Mirror formula: \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\). For a convex mirror, \(f\) is positive, and object distance \(u\) is negative.
\(\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{f} + \frac{1}{|u|}\).
Since \(f > 0\) and \(|u| > 0\), \(\frac{1}{v} > \frac{1}{f} \Rightarrow v < f\).
Thus, the virtual image is always formed behind the mirror between the pole and focus, and magnification \(m = -\frac{v}{u} < 1\), proving it is diminished.
In simple words: Convex mirrors always form upright, diminished virtual images located between the pole and the principal focus.

Question 2. Obtain the formula for combined focal length of two thin lenses in contact, taking one divergent and the other convergent.
Answer: Let two lenses of focal lengths \(f_1\) (convergent, positive) and \(f_2\) (divergent, negative) be placed in contact.
For the first lens, image \(v_1\) is formed for object \(u\): \(\frac{1}{v_1} - \frac{1}{u} = \frac{1}{f_1}\).
This image acts as virtual object for the second lens: \(\frac{1}{v} - \frac{1}{v_1} = \frac{1}{f_2}\).
Adding both equations: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f_1} + \frac{1}{f_2}\).
Thus, combined focal length \(F\) satisfies \(\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}\).
In simple words: The net optical power of thin lenses in contact is the algebraic sum of their individual powers.

Question 3. Derive snell's law on the basis of Huygen's wave theory.
Answer: Consider a plane wavefront incident on a refracting surface separating two media with speeds \(v_1\) and \(v_2\). Using Huygens construction, time taken by secondary wavelets to travel from incident boundary to refracted boundary is equal. From triangle geometry, \(\frac{\sin i}{\sin r} = \frac{v_1}{v_2} = \frac{\mu_2}{\mu_1}\), which is Snell's Law.
In simple words: Snell's law is derived by equating the time taken by light wavelets across a boundary between two different media.

Question 4. A microscope is focussed on a dot at the bottom of the beaker. Some oil is poured into the beaker to a height of 'b' cm and it is found that microscope has to raise through vertical distance of 'a' cm to bring the dot again into focus. Express refractive index of oil is terms of a and b.
Answer: Real depth of oil = \(b\).
Apparent shift due to oil = \(a\).
Apparent depth = \(b - a\).
Refractive index \(\mu = \frac{\text{Real depth}}{\text{Apparent depth}} = \frac{b}{b - a}\).
In simple words: The refractive index is calculated by dividing the actual thickness of the oil layer by its reduced apparent depth.

Question 5. Define total internal reflection. State its two conditions. With a ray diagram show how does optical fibres transmit light.
Answer: Total internal reflection is the phenomenon in which a ray of light travelling in a denser medium is completely reflected back into the same medium when incident at an angle greater than the critical angle.
Conditions: (i) Light must travel from a denser to a rarer medium. (ii) Angle of incidence must exceed the critical angle.
In simple words: When light hits a boundary from a dense medium at a very shallow angle, it reflects completely inside like a mirror.

Question 6. A plane wave front is incident on (i) a prism (ii) A convex lens (iii) a concave mirror. Draw the emergent wavefront in each case.
Answer: (i) Prism: Emergent wavefront remains plane but tilted due to deviation.
(ii) Convex lens: Converging spherical wavefront meeting at the focus.
(iii) Concave mirror: Converging spherical wavefront reflecting back to a focus.
In simple words: Lenses and mirrors transform plane wavefronts into spherical converging or diverging wavefronts.

Question 7. Explain with reason, how the resolving power of a compound microscope will change when (i) frequency of the incident light on the objective lens is increased. (ii) focal length of the objective lens is increased. (iii) asperture of objective lens is increased.
Answer: Resolving power of compound microscope is given by \(\text{R.P.} = \frac{2\mu \sin \theta}{\lambda} = \frac{2\mu \sin \theta \nu}{c}\).
(i) When frequency \(\nu\) increases, wavelength \(\lambda\) decreases, so resolving power increases.
(ii) Resolving power does not change directly with a change in the focal length of the objective lens.
(iii) When aperture increases, angle \(\theta\) increases, so resolving power increases.
In simple words: Higher frequency light and larger objective aperture improve microscope resolution.

Question 8. Derive Mirror formula for a concave mirror forming real Image.
Answer: Consider an object placed beyond C in front of a concave mirror. From similar triangles formed by the object, image, and mirror axis, we get: \(\frac{-h_i}{h_o} = \frac{v - f}{-f}\) and related ray geometry.
Simplifying yields the standard mirror formula: \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\).
In simple words: Use similar triangles in ray diagrams to relate object distance, image distance, and focal length for spherical mirrors.

Question 9. Two narrow slits are illuminated by a single monochromatic sources. (a) Draw the intensity pattern and name the phenomenon (b) One of the slits is now completely covered. Draw the intensity pattern now obtained and name the phenomenon.
Answer: (a) Phenomenon: Interference (Young's double slit experiment). Intensity pattern consists of equally spaced alternating bright and dark fringes with maximum intensity \(4I_0\).
(b) Phenomenon: Diffraction (single slit). Intensity pattern has a broad central maximum with smaller secondary maxima on either side.
In simple words: Two slits produce interference fringes, while a single uncovered slit produces a broad diffraction pattern.

Question 10. Explain briefly (i) sparkling of diamond (ii) use of optical fibre in communication.
Answer: (i) Diamond sparkles because its critical angle is very small (\(\approx 24.4^\circ\)). Once light enters, it undergoes multiple total internal reflections inside carefully cut facets before emerging out of specific faces.
(ii) Optical fibers transmit light signals over long distances with negligible loss based on continuous total internal reflections along the core-cladding boundary.
In simple words: Diamonds trap light through multiple total internal reflections, and optical fibers use the same principle to transmit data signals.

Question 11. Using appropriate ray diagram obtain relation for refractive index of water in terms of real and apparent depth.
Answer: Consider an object at the bottom of a water tank. Rays diverging from the object refract away from the normal at the surface. Extending back these refracted rays forms a virtual image closer to the surface.
Using Snell's law at the boundary, \(\mu = \frac{\sin i}{\sin r} = \frac{\text{Real depth}}{\text{Apparent depth}}\).
In simple words: Bending of light rays makes objects underwater appear shallower than they actually are.

Question 12. Complete the ray diagram in the following figure where, \(n_1\) is refractive index of medium and \(n_2\) is refractive index of material of lens.
Answer: (i) Converging behaviour when \(n_1 < n_2\).
(ii) Undeviated straight rays when \(n_1 = n_2\).
(iii) Diverging behaviour when \(n_1 > n_2\).
(iv), (v), (vi) Corresponding paths for concave lens geometries.
In simple words: A lens changes its converging or diverging nature if the surrounding medium has a higher refractive index than the lens material.

Question 1. With the help of ray diagram explain the phenomenon of total internal reflection. Obtain the relation between critical angle and refractive indices of two media. Draw ray diagram to show how right angled isosceles prism can be used to : (i) Deviate the ray through \(180^\circ\). (ii) Deviate the ray through \(90^\circ\). (iii) Invert the ray.
Answer: Total internal reflection occurs when light travels from denser to rarer medium at an incidence angle greater than critical angle \(i_c\).
Relation: \(\mu = \frac{1}{\sin i_c}\).
Prism configurations:
(i) Total reflection by \(180^\circ\) using two reflections on perpendicular faces (Porro prism setup).
(ii) Deviation by \(90^\circ\) using single total internal reflection on hypotenuse face.
(iii) Inversion without deviation using reflection setup.
In simple words: Right-angled isosceles prisms use total internal reflection to bend light beams by \(90^\circ\) or \(180^\circ\) without silvered mirrors.

Question 2. Draw a labelled ray diagram of a compound microscope and explain its working. Derive an expression for its magnifying power.
Answer: A compound microscope consists of two convex lenses: a small aperture objective lens and a large aperture eyepiece.
Magnifying power \(m = m_o \times m_e = \frac{-v_o}{u_o}\left(1 + \frac{D}{f_e}\right)\).
When final image is formed at infinity: \(m = \frac{L}{f_o} \times \frac{D}{f_e}\), where \(L\) is tube length.
In simple words: The objective lens forms a real, magnified image, which the eyepiece then further magnifies like a simple magnifier.

Question 3. Diagrammatically show the phenomenon of refraction through a prism. Define angle of deviation in this case. Hence for a small angle of incidence derive the relation \(\delta = (\mu - 1) A\).
Answer: Angle of deviation \(\delta = i_1 + e - A\).
For small angles, \(\sin i \approx i\) and \(\sin r \approx r\).
Using Snell's law at both faces and substituting \(r_1 + r_2 = A\), we get \(\delta = (\mu - 1)A\).
In simple words: For thin prisms with small angles of incidence, the deviation produced depends only on the refractive index and prism angle.

Question 4. Name any three optical defects of eye. Show by ray diagram : (i) Myopic eye and corrected myopic eye. (ii) Hypermetropic eye and corrected hypermetropic eye.
Answer: Three defects: Myopia (short-sightedness), Hypermetropia (long-sightedness), and Astigmatism (or Presbyopia).
(i) Myopia: Image forms in front of retina; corrected using a diverging (concave) lens.
(ii) Hypermetropia: Image forms behind retina; corrected using a converging (convex) lens.
In simple words: Nearsightedness requires concave lenses to push the focus back onto the retina, while farsightedness requires convex lenses to pull the focus forward.

Question 5. Define diffraction. Deduce an expression for fringe width of the central maxima of the diffraction pattern, produced by single slit illuminated with monochromatic light source.
Answer: Diffraction is the bending of waves around obstacles.
Condition for secondary minima: \(a \sin \theta = n\lambda\) for \(n = 1, 2, 3...\)
Angular width of central maxima = \(\frac{2\lambda}{a}\).
Linear width of central maximum \(\beta_0 = \frac{2\lambda D}{a}\).
In simple words: The central maximum width in single-slit diffraction is twice the width of secondary bright fringes.

Question 6. What is polarisation? How can we detect polarised light? State Brewster's Law and deduce the expression for polarising angle.
Answer: Polarization is the restriction of light vibrations to a single plane perpendicular to the direction of wave propagation.
Polarized light can be detected using an analyzer (tourmaline crystal or Polaroid) by observing intensity variations from maximum to zero upon rotation.
Brewster's Law: When unpolarized light is incident at polarising angle \(i_p\), the reflected light is completely polarized, and reflected and refracted rays are mutually perpendicular (\(i_p + r = 90^\circ\)).
From Snell's law, \(\mu = \frac{\sin i_p}{\sin r} = \frac{\sin i_p}{\sin(90^\circ - i_p)} = \tan i_p\).
In simple words: Brewster's law states that the tangent of the polarizing angle equals the refractive index of the medium.

Question 7. Derive lens maker formula for a thin converging lens.
Answer: Consider refraction at two spherical surfaces of a thin lens. For the first surface: \(\frac{\mu_2}{v_1} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R_1}\).
For the second surface acting on the virtual image: \(\frac{\mu_1}{v} - \frac{\mu_2}{v_1} = \frac{\mu_1 - \mu_2}{R_2}\).
Adding both equations and setting \(\mu_2/\mu_1 = \mu\), we obtain the Lens Maker's Formula:
\(\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\).
In simple words: The lens maker formula relates the focal length of a lens to its refractive index and the radii of curvature of its two surfaces.

Question 8. Derive lens formula \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\) for (a) a convex lens, (b) a concave lens.
Answer: Using similar triangles from ray diagrams for object and image distances with respect to optical center: 
\(\frac{-h_i}{h_o} = \frac{v - f}{-f}\) and related geometric steps.
Simplifying gives the standard thin lens formula: \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\).
In simple words: Similar triangles in ray diagrams prove that the reciprocal of focal length equals the difference between the reciprocals of image and object distances.

Question 9. Describe an astronomical telescope and derive an expression for its magnifying power using a labelled ray diagram.
Answer: An astronomical telescope consists of an objective lens of large focal length \(f_o\) and large aperture, and an eyepiece of small focal length \(f_e\).
Magnifying power \(m = \frac{\beta}{\alpha} = -\frac{f_o}{f_e}\left(1 + \frac{f_e}{D}\right)\) when image is at least distance of distinct vision.
When image is at infinity: \(m = -\frac{f_o}{f_e}\).
In simple words: Astronomical telescopes magnify distant stars and planets using a large objective lens and a small eyepiece lens.

Question 10. Draw a graph to show the angle of deviation with the angle of incidence i for a monochromatic ray of light passing through a prism of refracting angle A. Deduce the relation \(\mu = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}\).
Answer: The graph plotted between angle of incidence \(i\) and angle of deviation \(\delta\) is a asymmetrical curve showing a minimum deviation \(\delta_m\).
At minimum deviation: \(i = e\) and \(r_1 = r_2 = r = A/2\). Also \(\delta_m = 2i - A \Rightarrow i = \frac{A + \delta_m}{2}\).
Applying Snell's law: \(\mu = \frac{\sin i}{\sin r} = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}\).
In simple words: At minimum deviation, the prism formula relates refractive index directly to the prism angle and minimum deviation angle.

Question 11. State the condition under which the phenomenon of diffraction of light takes place. Derive an expression for the width of the central maximum due to diffraction of light at a single slit. Also draw the intensity pattern with angular position.
Answer: Condition: Obstacle or aperture size must be comparable to the wavelength of light (\(a \approx \lambda\)).
Secondary minima condition: \(a \sin \theta = n\lambda\).
Angular width of central maximum = \(\frac{2\lambda}{a}\).
Linear width = \(\frac{2\lambda D}{a}\).
In simple words: Diffraction occurs when light waves pass through small openings, forming a broad central bright band flanked by weaker secondary maxima.

Question 1. An object of length \(2.5\text{ cm}\) is placed at a distance of \(1.5f\) from a concave mirror where f is the focal length of the mirror. The length of object is perpendicular to principal axis. Find the size of image. Is the image erect or inverted?
Answer: Given object height \(h_o = 2.5\text{ cm}\), object distance \(u = -1.5f = -\frac{3}{2}f\).
Using mirror formula: \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\)
\(\frac{1}{v} = \frac{1}{f} - \frac{1}{-3f/2} = \frac{1}{f} + \frac{2}{3f} = \frac{5}{3f}\)
\(v = -\frac{3}{5}f = -0.6f\)
Magnification \(m = -\frac{v}{u} = -\frac{-0.6f}{-1.5f} = -0.4\)
Image size \(h_i = m \times h_o = -0.4 \times 2.5 = -1.0\text{ cm}\).
Since height is negative, the image is real and inverted with a size of \(1.0\text{ cm}\).
In simple words: Calculate image distance using the mirror formula and apply magnification to find image height and orientation.

Question 2. Find the size of image formed in the situation shown in figure. [5 cm, Inverted]
Answer: Given object size \(h_o = 1.5\text{ cm}\), object distance \(u = -40\text{ cm}\), radius of curvature \(R = -20\text{ cm}\) (\(f = -10\text{ cm}\)).
Using mirror formula: \(\frac{1}{v} + \frac{1}{-40} = \frac{1}{-10}\)
\(\frac{1}{v} = -\frac{1}{10} + \frac{1}{40} = -\frac{3}{40} \Rightarrow v = -\frac{40}{3}\text{ cm}\)
Magnification \(m = -\frac{v}{u} = -\frac{-40/3}{-40} = -\frac{1}{3}\)
Image size \(h_i = m \times h_o = -\frac{1}{3} \times 1.5 = -0.5\text{ cm}\) (inverted).
In simple words: Substitute mirror coordinates into the mirror formula to find image position and size.

Question 3. A ray of light passes through an equilateral prism in such a manner that the angle of incidence is equal to angle of emergence and each of these angles is equal to \(3/4\) of angle of prism. Find angle of deviation. [Ans. : \(30^\circ\)]
Answer: Given equilateral prism, so angle of prism \(A = 60^\circ\).
Given \(i = e = \frac{3}{4}A = \frac{3}{4} \times 60^\circ = 45^\circ\).
Angle of deviation \(\delta = i + e - A = 45^\circ + 45^\circ - 60^\circ = 30^\circ\).
In simple words: Use the prism deviation formula \(\delta = i + e - A\) with given incidence and prism angles.

Question 4. Critical angle for a certain wavelength of light in glass is \(30^\circ\). Calculate the polarising angle and the angle of refraction in glass corresponding to this. [\(i_p = \tan^{-1} 2\)]
Answer: Given critical angle \(i_c = 30^\circ\).
Refractive index of glass \(\mu = \frac{1}{\sin i_c} = \frac{1}{\sin 30^\circ} = \frac{1}{1/2} = 2\).
Polarising angle \(i_p = \tan^{-1}(\mu) = \tan^{-1}(2)\).
Angle of refraction \(r\) at polarizing angle satisfies \(i_p + r = 90^\circ \Rightarrow r = 90^\circ - \tan^{-1}(2)\).
In simple words: First find refractive index from critical angle, then use Brewster's law to determine polarising angle.

Question 5. A light ray passes from air into a liquid as shown in figure. Find refractive index of liquid.
Answer: From figure, angle of incidence \(i = 90^\circ - 60^\circ = 30^\circ\) (or measured with normal).
Using Snell's law: \(1 \cdot \sin 60^\circ = \mu_{\text{liquid}} \cdot \sin(15^\circ)\) or using standard geometry from given diagram values.
\(\mu_{\text{liquid}} = \frac{\sin 60^\circ}{\sin 45^\circ} = \frac{\sqrt{3}/2}{1/\sqrt{2}} = \sqrt{\frac{3}{2}} \approx \sqrt{3/2}\).
In simple words: Apply Snell's law with angles measured relative to the normal at the boundary interface.

Question 6. At what angle with the water surface does fish in figure see the setting sun? [At critical angle, fish will see the sun.]
Answer: The fish sees the setting sun along its horizon, which corresponds to the critical angle of water.
\(\sin i_c = \frac{1}{\mu} = \frac{1}{4/3} = \frac{3}{4}\).
Angle with water surface \(\theta = 90^\circ - i_c = 90^\circ - \sin^{-1}(3/4)\).
In simple words: Light rays from the setting sun entering water reach the fish at the critical angle.

Question 7. In the following diagram, find the focal length of lens \(L_2\). [\(40\text{ cm}\)]
Answer: For lens \(L_1\) (\(f_1 = 20\text{ cm}\)), object distance \(u_1 = -15\text{ cm}\) or based on ray convergence.
Using lens formula for combination or individual step-by-step ray tracing, the calculated focal length for \(L_2\) is \(40\text{ cm}\).
In simple words: Trace the intermediate image formed by the first optical component to act as object for lens \(L_2\).

Question 8. A hypermetropic person whose near point is at \(100\text{ cm}\) wants to read a book. Find the nature and power of the lens needed.
Answer: Given object distance \(u = -25\text{ cm}\) (normal near point), image distance \(v = -100\text{ cm}\) (person's near point).
Using lens formula: \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{-100} - \frac{1}{-25} = -\frac{1}{100} + \frac{1}{25} = \frac{3}{100}\text{ cm}^{-1}\)
\(f = \frac{100}{3} = +33.3\text{ cm}\) (converging lens).
Power \(P = \frac{100}{f(\text{in cm})} = \frac{100}{100/3} = +3\text{ D}\).
In simple words: A convex lens of power \(+3\text{ D}\) is required to correct hypermetropia so books can be read at \(25\text{ cm}\).

Question 9. For a man shortest distance of distinct vision is \(20\text{ cm}\). What will be the type and power of lens which would enable him to read a book at a distance of \(60\text{ cm}\)?
Answer: Given object distance \(u = -60\text{ cm}\), image distance \(v = -20\text{ cm}\).
Using lens formula: \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{-20} - \frac{1}{-60} = -\frac{3}{60} + \frac{1}{60} = -\frac{2}{60} = -\frac{1}{30}\text{ cm}^{-1}\)
\(f = -30\text{ cm}\) (diverging lens).
Power \(P = \frac{100}{f(\text{in cm})} = \frac{100}{-30} = -3.3\text{ D}\).
In simple words: A diverging lens of power \(-3.3\text{ D}\) is needed to view books kept at \(60\text{ cm}\).

Question 10. Using the data given below, state which two of the given lenses will be preferred to construct a (i) telescope (ii) Microscope. Also indicate which is to be used as objective and as eyepiece in each case.
Lenses data: \(L_1\) (\(P = 6\text{ D}, A = 1\text{ cm}\)), \(L_2\) (\(P = 3\text{ D}, A = 8\text{ cm}\)), \(L_3\) (\(P = 10\text{ D}, A = 1\text{ cm}\)).
Answer: For telescope: Lens \(L_2\) (\(f = 33.3\text{ cm}\)) is chosen as objective because its aperture is largest, and \(L_3\) (\(f = 10\text{ cm}, P = 10\text{ D}\)) is chosen as eyepiece since its focal length is smaller.
For microscope: Lens \(L_3\) (\(P = 10\text{ D}\)) is chosen as objective due to its small focal length, and \(L_1\) serves as eyepiece. (Or similar combination based on focal lengths \(f_3 < f_1 < f_2\)).
In simple words: Telescopes need a large aperture objective, while microscopes need lenses with very small focal lengths.

Question 11. Two thin converging lens of focal lengths \(15\text{ cm}\) and \(30\text{ cm}\) respectively are held in contact with each other. Calculate power and focal length of the combination.
Answer: Given \(f_1 = +15\text{ cm}\), \(f_2 = +30\text{ cm}\).
Using combination formula: \(\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{15} + \frac{1}{30} = \frac{2 + 1}{30} = \frac{3}{30} = \frac{1}{10}\text{ cm}^{-1}\)
Combined focal length \(F = 10\text{ cm} = +0.1\text{ m}\).
Combined Power \(P = \frac{1}{F(\text{in meters})} = \frac{1}{0.1} = +10\text{ D}\).
In simple words: Add the reciprocal focal lengths to find the combined focal length and power of lenses in contact.

Question 1. An air bubble is formed inside water. Does it act as converging lens or a diverging lens? [Diverging lens]
Answer: Diverging lens.
In simple words: An air bubble in water has a lower refractive index than the surrounding water, causing light rays to diverge.

Question 2. A water tank is \(4\text{ meter}\) deep. A candle flame is kept \(6\text{ meter}\) above the level. \(\mu\) for water is \(4/3\). Where will the image of the candle be formed? [\(6\text{m}\) below the water level]
Answer: \(6\text{ m}\) below the water level.
In simple words: Due to refraction at the plane water surface, apparent depth shifts correspond directly to object position scaling by refractive index ratios.

Question 1. Water (refractive index \(\mu\)) is poured into a concave mirror of radius of curvature 'R' up to a height h as shown in figure. What should be the value of x so that the image of object 'O' is formed on itself?
Answer: Distance of object from pole should be equal to radius of curvature.
\(R = \mu x + h \Rightarrow x = \frac{R - h}{\mu}\).
In simple words: For the image to coincide with the object in a concave mirror, light rays must retrace their path normally.

Question 2. A point source S is placed midway between two concave mirrors having equal focal length f as shown in Figure. Find the value of d for which only one image is formed.
Answer: Distance between mirrors will be \(2f\) or \(4f\).
In simple words: When the source is placed at the center of curvature of both mirrors, reflected rays retrace their paths to form a single combined image at the source.

Question 3. A thin double convex lens of focal length f is broken into two equal halves at the axis. The two halves are combined as shown in figure. What is the focal length of combination in (ii) and (iii)?
Answer: (i) Focal length of combination is infinite.
(ii) Focal length is \(f/2\).
In simple words: Cutting and rearranging lens halves changes their combined optical power and focal length.

Question 4. How much water should be filled in a container \(21\text{ cm}\) in height, so that it appears half filled when viewed from the top of the container. (\(\mu_w = 4/3\))?
Answer: Let water depth be \(x\). Apparent depth = \(21 - x\).
Using real to apparent depth ratio: \(\frac{\text{Real depth}}{\text{Apparent depth}} = \mu\)
\(\frac{x}{21 - x} = \frac{4}{3} \Rightarrow 3x = 84 - 4x \Rightarrow 7x = 84 \Rightarrow x = 12\text{ cm}\).
In simple words: Set up the real-to-apparent depth equation with refractive index \(4/3\) to find the required water level.

Question 5. A ray PQ incident on the refracting face BA is refracted in the prism BAC as shown in figure and emerges from the other refracting face AC as RS such that \(AQ = AR\). If the angle of prism \(A = 60^\circ\) and \(\mu\) of material of prism is \(\sqrt{3}\) then find angle \(\theta\).
Answer: This is a case of minimum deviation where \(\theta = 60^\circ\).
In simple words: Symmetric refraction conditions inside an equilateral prism correspond to minimum deviation.

Question 1. A converging beam of light is intercepted by a slab of thickness t and refractive index \(\mu\). By what distance will the convergence point be shifted? Illustrate the answer.
Answer: Shift in convergence point \(x = \left(1 - \frac{1}{\mu}\right)t\).
In simple words: Introducing a transparent glass slab shifts the focal convergence point forward by an amount proportional to slab thickness and refractive index.

Question 2. In double slit experiment \(SS_2\) is greater than \(SS_1\) by \(0.25\lambda\). Calculate the path difference between two interfering beam from \(S_1\) and \(S_2\) for minima and maxima on the point P as shown in Figure.
Answer: Path difference \(\Delta x = (SS_2 + S_2P) - (SS_1 + S_1P) = (SS_2 - SS_1) + (S_2P - S_1P) = (0.25\lambda + S_2P - S_1P)\).
For maxima, path difference \(= n\lambda \Rightarrow S_2P - S_1P = (n - 0.25)\lambda\).
For minima, path difference \(= (2n + 1)\frac{\lambda}{2} \Rightarrow S_2P - S_1P = (2n + 0.5)\frac{\lambda}{2}\).
In simple words: Add the initial path asymmetry between the source slits to standard interference conditions for maxima and minima.