CBSE Class 12 Physics Ray and Wave Optics Assignment Set 01

TEST SERIES(PHYSICS)

RAY AND WAVE OPTICS

Question. In Young's double slit experiment with sodium vapour lamp of wavelength 589 nm and the slits 0.589 mm apart, the half angular width of the central maximum is
(a) sin^–1 0.01
(b) sin^–10.0001
(c) sin^–1 0.001
(d) sin^–1 0.1

Answer: C

Question. The separation between successive fringes in a double slit arrangement is x. If the whole arrangement is dipped under water what will be the new fringe separation? [The wavelenght of light being used is 5000 Å]
(a) 1.5 x
(b) x
(c) 0.75 x
(d) 2 x

Answer: C

Question. When the angle of incidence is 60° on the surface of a glass slab, it is found that the reflected ray is completely polarised. The velocity of light in glass is
(a) √2 x 10⁸ms^-1
(b) √3 x 10⁸ms^-1
(c) 2 x 10⁸ms^-1
(d) 3 x 10⁸ms-1

Answer: B

Question. Young's double slit experment is first performed in air and then in a medium other than air. It is found that 8th bright fringe in the medium lies where 5th dark fringe lies in air. The refractive index of the medium is nearly 
(a) 1.59
(b) 1.69
(c) 1.78
(d) 1.25

Answer: C

Question. A ray of light is incident on the surface of a glass plate at an angle of incidence equal to Brewster’s angel f. If m represents the refractive index of glass with respect to air, then the angle between the reflected and the refracted rays is
(a) 90° + Φ
(b) sin^–1(m cos Φ)
(c) 90º
(d) 90º sin^-1(sin Φ/μ)

Answer: C

Question. A rocket is receding away from earth with velocity 0.2 c. The rocket emits signal of frequency 4 × 10⁷ Hz. The apparent frequency observed by the observer on earth will be
(a) 4 × 10⁶ Hz
(b) 3.2 × 10⁷ Hz
(c) 3 × 10⁶ Hz
(d) 5 × 10⁷ Hz

Answer: B

Question. Consider a ray of light incident from air onto a slab of glass (refractive index n) of width d, at an angle q. The phase difference between the ray reflected bv the top surface of the glass and the bottom surface is 

Answer: A

Question. At the first minimum adjacent to the central maximum of a single-slit diffraction pattern, the phase difference between the Huygen's wavelet from the edge of the slit and the wavelet from the midpoint of the slit is :
(a) π/2 radian
(b) π radian 
(c) π/8 radian
(d) π/4 radian

Answer: B

Question. The wavefronts of a light wave travelling in vacuum are given by x + y + z = c. The angle made by the direction of propagation of light with the X-axis is
(a) 0º
(b) 45º
(c) 90º
(d) cos^–1(1/√3)

Answer: D

Question. The width of a slit is 0.012 mm. Monochromatic light is incident on it. The angular position of first bright line is 5.2º. The wavelength of incident light is
[sin 5.2º = 0.0906].
(a) 6040 Å
(b) 4026 Å
(c) 5890 Å
(d) 7248 Å

Answer: D

Question. In Fresnel’s biprism expt., a mica sheet of refractive index 1.5 and thickness 6 × 10^–6 m is placed in the path of one of interfering beams as a result of which the central fringe gets shifted through 5 fringe widths. The wavelength of light used is
(a) 6000 Å
(b) 8000 Å
(c) 4000 Å
(d) 2000 Å

Answer: A

Question. In young’s double-slit experiment, the intensity of light at a point on the screen where the path difference is λ is I, λ being the wavelength of light used. The intensity at a point where the path difference is λ/4 will be
(a) I/4
(b) I/2
(c) I
(d) zero

Answer: B

Question. In Young's double slit experiment, the fringes are displaced by a distance x when a glass plate of refractive index 1.5 is introduced in the path of one of the beams. When this plate is replaced by another plate of the same thickness, the shift of fringes is (3/2) x. The refractive index of the second plate is
(a) 1.75
(b) 1.50
(c) 1.25
(d) 1.00

Answer: A

Question. Light of wavelength 6.5 × 10^–7 m is made incident on two slits 1 mm apart. The distance between third dark fringe and fifth bright fringe on a screen distant 1 m from the slits will be
(a) 0.325 mm
(b) 0.65 mm
(c) 1.625 mm
(d) 3.25 mm

Answer: C

Question. Two nicols are oriented with their principal planes making an angle of 60º. Then the percentage of incident unpolarised light which passes through the system is
(a) 100
(b) 50
(c) 12.5
(d) 37.5

Answer: C

Question. When the light is incident at the polarizing angle on the transparent medium, then the completely polarized light is
(a) refracted light
(b) reflected light
(c) refracted and reflected light
(d) neither reflected nor refracted light

Answer: B

Question. The ratio of resQuestionolving powers of an optical microscope for two wavelengths λ₁ = 4000 Å and λ₂ = 6000 Å is 
(a) 9 : 4
(b) 3 : 2
(c) 16 : 81
(d) 8 : 27

Answer: B

Question. A beam of unpolarised light passes through a tourmaline crystal A and then through another such crystal B oriented so that the principal plane is parallel to A. The intensity of emergent light is I0. Now B is rotated by 45º about the ray.
The emergent light will have intensity
(a) I₀ / 2
(b) I₀ /√2
(c) I₀√2
(d) 2I₀

Answer: A

Question. In a double slit experiment, the two slits are 1 mm apart and the screen is placed 1 m away. A monochromatic light wavelength 500 nm is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of single slit pattern ? 
(a) 0.1 mm
(b) 0.5 mm
(c) 0.02 mm
(d) 0.2 mm

Answer: D

Question. A parallel beam of fast moving electrons is incident normally on a narrow slit. A fluorescent screen is placed at a large distance from the slit. If the speed of the electrons is increased, which of the following statements is correct ?
(a) The angular width of the central maximum of the diffraction pattern will increase.
(b) The angular width of the central maximum will decrease.
(c) The angular width of the central maximum will be unaffected.
(d) Diffraction pattern is not observed on the screen in case of electrons.

Answer: B

THREE MARKS QUESTION

Question. Draw the type of wave front that corresponds to a beam of monochromatic light
(i) Coming from a very far off source and
(ii) Diverging radially from a point source
Answer : (i) Plane wave front (ii) Spherical wavefront

Question. Sketch the wave fronts corresponding to
(a) Converging rays
(b) Diverging rays
Answer : (i) wavefront is convex (ii) wavefront is concave

Question. In young’s double slit experiment, explain with reason in each case, how the interference pattern changes, when
(i) Separation between the slits is increased and
(ii) Screen is moved away from the plane of the slits.
Answer : . β = Dλ⁄d(i) Fringe width decreases as it is inversely proportional to the distance between the slits.
(ii) Fringe width is increased as it is directly proportional to the distance between slits and screen.

Question. How will the angular separation and visibility of fringes in Young’s double slit experiment change when
(i) Screen is moved away from the plane of the slits?
(ii) Wavelength of light is increased?
Answer : θ = λ⁄d (i) there will be no effect on angular fringe width.
(ii) Angular separation increases as it is directly proportional to the wavelength.

Question. In what way is diffraction from each slit is related to the interference pattern in a double slit experiment?
Answer : In Young’s double slit experiment there will be two diffraction pattern from two slits after superpose
it results into interference pattern.

Question. Define resolving power of a compound microscope. How does the resolving power of a compound microscope changes when –
(i) Refractive index of medium between the object and the objective lens increases.
(ii) Wavelength of radiation used is increased.
(iii) Decreasing the diameter of its objective.
Answer : Definition of resolving power. (i) R P increases (ii) RP decreases (iii) RP decreases

Question. How would the diffraction pattern of a single slit be affected when:
(i) The width of the slit is decreased
(ii) The monochromatic light is replaced with a light of higher frequency
(iii) The separation between the slit and screen is increased.
Answer : (i) Area of illumination increases
(ii) Fringe width is decreases
(iii) Fringe width increases

Question. (a) Using the phenomena of polarization, show how transverse nature of light can be demonstrated.
(b) Two polaroids P1 and P2 are placed with their axes perpendicular to each other. Unpolarised light of intensity I0 is incident on P1. A third Polaroid P3 is kept between P1 and P2 such that its axis makes an angle of 300 with that of P`1. Determine the intensity of light transmitted through P1 , P2 and P3.
Answer : When light is passed through a polarizer followed by an analyser, the intensity variation on the screen shows that light is transverse in nature.
Using the formula I = I0cos2Θ:
Θ1 = 300 angle between P1 and P3 and Θ2 = 600 angle between P3 and P2
The final intensity is 3I0/32

Question. (a) What is linearly polarized light? Describe briefly using a diagram how sunlight is polarized?
(b) Unpolarized light is incident on a Polaroid. How would the intensity of transmitted light change when the Polaroid is rotated?
Answer : (a) Definition of linearly polarized. Sunlight is polarized by scattering of light.
(b) there is no change in intensity of light.

Question. Two polaroids are placed at 900 to each other and transmitted intensity is zero. What happens when one more polaroid is placed between these two bisecting the angle between them. Take intensity of unpolarised light is I0.
Answer : Using the formula I = I0cos2Θ:
Θ1 = 450 angle between P1 and P3 and Θ2 = 450 angle between P3 and P2.
The final intensity is I0/4

Important Questions for NCERT Class 12 Physics Wave Optics

Question. Which one of the following phenomena is not explained by Huygen’s construction of wavefront ?
(a) Refraction
(b) Reflection
(c) Diffraction
(d) Origin of spectra

Answer : D

Question. The frequency of a light wave in a material is 2 × 10¹⁴ Hz and wavelength is 5000 Å. The refractive index of material will be
(a) 1.50
(b) 3.00
(c) 1.33
(d) 1.40

Answer : B

Question. In Young’s double slit experiment, we get 10 fringes in the field of view of monochromatic light of wavelength 4000Å. If we use monochromatic light of wavelength 5000Å, then the number of fringes obtained in the same field of view is 
(a) 8
(b) 10
(c) 40
(d) 50

Answer : A

Question. The Fraunhoffer ‘diffraction’ pattern of a single slit is formed in the focal plane of a lens of focal length 1 m. The width of slit is 0.3 mm. If third minimum is formed at a distance of 5 mm from central maximum, then wavelength of light will be 
(a) 5000 Å
(b) 2500 Å
(c) 7500 Å
(d) 8500 Å

Answer : A

Question. The refractive index of water is 1.33. What will be the speed of light in water?
(a) 4 × 10⁸ m/s
(b) 1.33 × 10⁸ m/s
(c) 3 × 10⁸ m/s
(d) 2.25 × 10⁸ m/s

Answer : D

Question. Light travels through a glass plate of thickness t and having a refractive index m. If c is the velocity of light in vacuum, the time taken by light to travel this thickness of glass is
(a) t/μc
(b) μt/c
(c) tmc
(d) tc/μ

Answer : B

Question. A star, which is emitting radiation at a wavelength of 5000 Å, is approaching the earth with a velocity of 1.5 × 10⁴ m/s. The change in wavelength of the radiation as received on the earth is
(a) 25 Å
(b) 100 Å
(c) zero
(d) 2.5 Å 

Answer : A

Question. Time taken by sunlight to pass through a window of thickness 4 mm whose refractive index is 3/2 is
(a) 2 × 10^–4 s
(b) 2 × 10⁸ s
(c) 2 × 10^–11 s
(d) 2 × 10¹¹ s 

Answer : C

Question. A beam of monochromatic light is refracted from vacuum into a medium of refractive index 1.5. The wavelength of refracted light will be
(a) depend on intensity of refracted light
(b) same
(c) smaller
(d) larger. 

Answer : C

Question. Green light of wavelength 5460 Å is incident on an air-glass interface. If the refractive index of glass is 1.5, the wavelength of light in glass would be (c = 3 × 10⁸ m s^–1)
(a) 3640 Å
(b) 5460 Å
(c) 4861 Å
(d) none of these.

Answer : A

Question. Ratio of intensities of two waves are given by 4 : 1. Then ratio of the amplitudes of the two waves is
(a) 2 : 1
(b) 1 : 2
(c) 4 : 1
(d) 1 : 4 

Answer : A

Question. Interference is possible in
(a) light waves only
(b) sound waves only
(c) both light and sound waves
(d) neither light nor sound waves

Answer : C

Question. In Young’s double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes
(a) double
(b) half
(c) four times
(d) one-fourth

Answer : C

Question. In a double slit experiment, when light of wavelength 400 nm was used, the angular width of the first minima formed on a screen placed 1 m away, was found to be 0.2°. What will be the angular width of the first minima, if the entire experimental apparatus is immersed in water? (mwater = 4/3)
(a) 0.1°
(b) 0.266°
(c) 0.15°
(d) 0.05°

Answer : C

Question. In a Young’ double slit experiment if there is no initial phase difference between the light from the two slits, a point on the screen corresponding to the fifth minimum has path difference.
(a) 5.λ/2
(b) 10.λ/2
(c) 9.λ/2
(d) 11.λ/2

Answer : C

Question. In Young’s double slit experiment the separation d between the slits is 2 mm, the wavelength l of the light used is 5896 Å and distance D between the screen and slits is 100 cm. It is found that the angular width of the fringes is 0.20°. To increase the fringe angular width to 0.21° (with same l and D) the separation between the slits needs to be changed to
(a) 1.8 mm
(b) 1.9 mm
(c) 2.1 mm
(d) 1.7 mm 

Answer : B

Question. Young’s double slit experiment is first performed in air and then in a medium other than air. It is found that 8th bright fringe in the medium lies where 5th dark fringe lies in air. The refractive index of the medium is nearly
(a) 1.59
(b) 1.69
(c) 1.78
(d) 1.25

Answer : C

Question. The intensity at the maximum in a Young’s double slit experiment is I0. Distance between two slits is d = 5l, where l is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance D = 10d ?
(a) 3/4 I₀
(b) I₀/2
(c) I₀
(d) I_0^/ 4

Answer : B

Question. Two slits in Young’s experiment have widths in the ratio 1 : 25. The ratio of intensity at the maxima and minima in the interference pattern, Imax /I min is
(a) 49/121
(b) 4/9
(c) 9/4
(d) 121/49

Answer : C

Question. A single slit Fraunhoffer diffraction pattern is formed with white light. For what wavelength of light the third secondary maximum in the diffraction pattern coincides with the second secondary maximum in the pattern for red light of wavelength 6500 Å? 
(a) 4400 Å
(b) 4100 Å
(c) 4642.8 Å
(d) 9100 Å

Answer : C

Question. At the first minimum adjacent to the central maximum of a single-slit diffraction pattern, the phase difference between the Huygen’s wavelet from the edge of the slit and the wavelet from the midpoint of the slit is 
(a) π radian
(b) π /8 radian
(c) π/4 radian
(d) π/2 radian 

Answer: A

Question. A beam of light of l = 600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between first dark fringes on either side of the central bright fringe is 
(a) 1.2 cm
(b) 1.2 mm
(c) 2.4 cm
(d) 2.4 mm 

Answer: D

Question. A parallel beam of fast moving electrons is incident normally on a narrow slit. A fluorescent screen is placed at a large distance from the slit. If the speed of the electrons is increased, which of the following statements is correct? 
(a) The angular width of the central maximum will decrease.
(b) The angular width of the central maximum will be unaffected.
(c) Diffraction pattern is not observed on the screen in the case of electrons.
(d) The angular width of the central maximum of the diffraction pattern will increase.

Answer: A
 
RAY AND WAVE OPTICS

Question 1. The intensity of the light transmitted by a polaroid is half the intensity of the light incident on it. Give reason.
Answer: Unpolarized light consists of electric field vibrations in all possible directions perpendicular to the direction of wave propagation, with equal probability in all directions.
When unpolarized light passes through a polaroid, only the electric field component parallel to the pass-axis of the polaroid is transmitted, while the component perpendicular to it is blocked.
If \(I_0\) is the intensity of the unpolarized incident light, the average value of \(\cos^2\theta\) over all directions is \(\frac{1}{2}\). Therefore, the transmitted intensity is given by:
\(I = \frac{I_0}{2}\)
In simple words: Unpolarized light vibrates in all directions, but a polaroid only allows vibrations along one specific axis to pass through, cutting the total light intensity exactly in half.

Exam Tip: Always state Malus law principles or average statistical components when explaining polaroid intensity reductions.

Question 2. Draw the shape of the refracted/reflected wave front when a plane wavefront incident on (i) prism (ii) convex mirror.
Answer: (i) When a plane wavefront is incident on a prism, the lower part of the wavefront travels through a greater thickness of glass than the upper part. Since the speed of light is less in glass than in air, the lower part is delayed more than the upper part, resulting in a tilted/deviated plane wavefront after refraction.
(ii) When a plane wavefront is incident on a convex mirror, the central part of the wavefront travels a lesser distance before reflection compared to the marginal parts. Consequently, the reflected wavefront converges virtually behind the mirror, forming a diverging spherical wavefront.
In simple words: A prism tilts a plane wavefront because different parts travel through different thicknesses, while a convex mirror reflects a plane wavefront into a diverging spherical wavefront.

Exam Tip: Clearly label incident wavefronts, reflected/refracted wavefronts, and ray directions in optical diagrams.

Question 3. Two objective lenses of aperture A1 and A2 (A1 > A2) are used in two telescope. Write the ratio of their resolving power.
Answer: The resolving power (RP) of an astronomical telescope is given by the formula:
\(\text{Resolving Power} = \frac{D}{1.22\lambda}\)
where \(D\) is the diameter (aperture) of the objective lens and \(\lambda\) is the wavelength of light used.
Since resolving power is directly proportional to the aperture diameter (\(\text{RP} \propto D\)), the ratio of the resolving powers of the two telescopes is:
\(\frac{(\text{RP})_1}{(\text{RP})_2} = \frac{A_1}{A_2}\)
In simple words: Resolving power depends directly on the lens aperture, so the ratio of resolving powers is simply equal to the ratio of their apertures, \(A_1 : A_2\).

Exam Tip: Remember that resolving power increases with larger aperture diameters to capture finer details of distant stars.

Question 4. For which angle of incidence a ray suffers minimum deviation of 30 degrees when incident on an equilateral prism.
Answer: For an equilateral prism, the angle of the prism is \(A = 60^\circ\).
Given the angle of minimum deviation \(\delta_m = 30^\circ\).
The angle of incidence \(i\) for minimum deviation is given by the formula:
\(i = \frac{A + \delta_m}{2}\)
Substituting the values:
\(i = \frac{60^\circ + 30^\circ}{2} = \frac{90^\circ}{2} = 45^\circ\)
In simple words: The angle of incidence for minimum deviation is calculated by adding the prism angle and minimum deviation, then dividing by 2.

Exam Tip: For minimum deviation in prisms, always use \(i = (A + \delta_m)/2\) and \(r = A/2\).

Question 5. A convex lens and a concave lens of same focal length are kept in contact. find the focal length of the combination.
Answer: Let the focal length of the convex lens be \(f_1 = +f\) and the focal length of the concave lens be \(f_2 = -f\).
The formula for the equivalent focal length \(F\) of two lenses in contact is:
\(\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}\)
Substituting the values:
\(\frac{1}{F} = \frac{1}{f} + \left(-\frac{1}{f}\right) = 0\)
\(F = \frac{1}{0} = \infty\)
The focal length of the combination is infinity, meaning the combination acts as a plane glass plate.
In simple words: Combining a converging convex lens and a diverging concave lens of identical focal lengths completely cancels out their optical powers, yielding an infinite focal length.

Question 6. On the basis of Huygens principle prove laws of reflection of light.
Answer: Consider a plane wavefront \(AB\) incident obliquely on a reflecting surface \(XY\) at an angle of incidence \(i\).
According to Huygens principle, every point on the wavefront acts as a source of secondary wavelets.
Let light travel from \(B\) to \(B'\) in time \(t\) with velocity \(v\), so distance \(BB' = vt\).
In the same time \(t\), secondary wavelets from point \(A\) travel a distance \(AA' = vt\) in the same medium.
Drawing a tangent \(A'B'\) from \(A'\) to the sphere of radius \(vt\) gives the reflected wavefront \(A'B'\).
In triangles \(\triangle ABB'\) and \(\triangle AA'B'\):
- \(BB' = AA' = vt\) (same speed and time)
- \(\angle ABB' = \angle AA'B' = 90^\circ\)
- \(AB'\) is common to both triangles.
Therefore, \(\triangle ABB' \cong \triangle AA'B'\) by RHS congruence.
From congruence, angle of incidence \(\angle BAB' = i\) is equal to angle of reflection \(\angle AB'A' = r\), proving that \(i = r\). Also, the incident ray, normal, and reflected ray all lie in the same plane, proving the laws of reflection.
In simple words: By constructing secondary wavelets using Huygens principle and proving triangle congruence, we show that the angle of incidence equals the angle of reflection.

Exam Tip: Clearly draw the incident wavefront, reflecting surface, secondary wavelets, and reflected wavefront to secure full marks.

Question 7. In Youngs double slit experiment in I is the intensity emitted by each slit then prove that in interference pattern intensity varies between 0 and 4I.
Answer: Let the amplitude of light wave from each individual slit be \(a\). The intensity of light is proportional to the square of the amplitude: \(I \propto a^2\).
When two waves interfere, the resultant amplitude \(R\) at any point is given by:
\(R^2 = a^2 + a^2 + 2aa\cos\phi\)
where \(\phi\) is the phase difference between the two interfering waves.
Since intensity \(I_{\text{res}} \propto R^2\), the resultant intensity is:
\(I_{\text{res}} = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi\)
Given \(I_1 = I_2 = I\):
\(I_{\text{res}} = I + I + 2\sqrt{I \cdot I}\cos\phi = 2I(1 + \cos\phi)\)
1. For constructive interference (bright fringe), phase difference \(\phi = 0, 2\pi, 4\pi, \dots\), so \(\cos\phi = 1\):
\(I_{\max} = 2I(1 + 1) = 4I\)
2. For destructive interference (dark fringe), phase difference \(\phi = \pi, 3\pi, 5\pi, \dots\), so \(\cos\phi = -1\):
\(I_{\min} = 2I(1 - 1) = 0\)
Thus, the intensity in the interference pattern varies between \(0\) and \(4I\).
In simple words: Maximum intensity occurs when waves add up in phase (\(4I\)), and minimum intensity occurs when they cancel out completely (\(0\)).

Exam Tip: Never add individual intensities directly (\(I + I = 2I\)). Always add amplitudes first, then square the result for interference.

Question 8. For which value of X first maxima will be obtained at P. Given \(\lambda = 500\text{ nm}\), angle \(30^\circ\).
Answer: In a single-slit diffraction or Young's double slit interference setup where a path difference is introduced by geometry, the condition for the first maxima (or path difference relation for constructive interference) is given by:
\(d \sin\theta = n\lambda\)
For the first maxima, \(n = 1\).
Here, the path difference is given by \(X \sin\theta\) or the geometric path length difference \(X\) corresponding to angle \(30^\circ\):
\(X \sin(30^\circ) = 1 \cdot \lambda\)
Given \(\lambda = 500\text{ nm} = 500 \times 10^{-9}\text{ m}\) and \(\theta = 30^\circ\):
\(X \cdot \sin(30^\circ) = 500\text{ nm}\)
\(X \cdot \frac{1}{2} = 500\text{ nm}\)
\(X = 2 \times 500\text{ nm} = 1000\text{ nm} = 1\text{ }\mu\text{m}\)
In simple words: Using the path difference formula \(X\sin\theta = \lambda\) with \(\theta = 30^\circ\) and \(\lambda = 500\text{ nm}\), we find distance \(X\) is \(1000\text{ nm}\).

Exam Tip: Always substitute standard SI units (meters) when calculating final distances unless asked otherwise.

Question 9. If an object is placed at the bottom of a tank of depth d. Refractive index of water is \(\mu\). Find the area of the circle within which we can see the object from the surface of water.
Answer: Light rays originating from an object at the bottom of a tank emerge into the air only if the angle of incidence is less than or equal to the critical angle \(C\).
Rays incident at an angle greater than \(C\) undergo total internal reflection and remain inside the water.
Thus, the visible light emerges through a circular patch of radius \(r\) on the water surface.
From Snell's law at the boundary, \(\sin C = \frac{1}{\mu}\).
From the geometry of the cone of light, if depth is \(d\) and radius of the circle is \(r\):
\(\tan C = \frac{r}{d} \implies r = d \tan C\)
Since \(\sin C = \frac{1}{\mu}\), we have \(\tan C = \frac{1}{\sqrt{\mu^2 - 1}}\).
Therefore, radius \(r = \frac{d}{\sqrt{\mu^2 - 1}}\).
The area \(A\) of the circle of visibility is:
\(A = \pi r^2 = \pi \left(\frac{d}{\sqrt{\mu^2 - 1}}\right)^2 = \frac{\pi d^2}{\mu^2 - 1}\)
In simple words: The visible circle is formed by light escaping within the critical angle cone. Using trigonometry and Snell's law, the area is \(\frac{\pi d^2}{\mu^2 - 1}\).

Exam Tip: Remember the critical angle relation \(\sin C = 1/\mu\) and derive the radius using \(r = d \tan C\).

Question 10. Focal length of double equi-convex lens is twice of radius of curvature of its each face. Find the refractive index of lens. What would be the nature of lens if it is immersed in a liquid of refractive index 2.
Answer: For a double equi-convex lens, let the radius of curvature of both faces be \(R_1 = +R\) and \(R_2 = -R\).
Given focal length \(f = 2R\).
Using the Lens Maker's Formula:
\(\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\)
\(\frac{1}{2R} = (\mu - 1)\left(\frac{1}{R} - \frac{1}{-R}\right) = (\mu - 1)\left(\frac{2}{R}\right)\)
Multiplying both sides by \(R\):
\(\frac{1}{2} = 2(\mu - 1)\)
\(\mu - 1 = \frac{1}{4} = 0.25 \implies \mu = 1.25\)
Now, if the lens is immersed in a liquid of refractive index \(\mu_l = 2\):
Using the relative lens maker's formula: 
\(\frac{1}{f_m} = \left(\frac{\mu}{\mu_l} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\)
\(\frac{1}{f_m} = \left(\frac{1.25}{2} - 1\right)\left(\frac{2}{R}\right) = (0.625 - 1)\left(\frac{2}{R}\right) = (-0.375)\left(\frac{2}{R}\right) = -\frac{0.75}{R}\)
Since the focal length \(f_m\) is negative, the lens behaves as a diverging (concave) lens when immersed in this liquid.
In simple words: The lens refractive index is 1.25. When placed in a liquid of higher refractive index (\(2\)), light bends outward, turning the convex lens into a diverging lens.

Exam Tip: A lens behaves as a diverging lens if placed in a surrounding medium whose refractive index is greater than that of the lens material.

Question 11. For concave spherical surface prove that \(\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}\).
Answer: Consider a point object \(O\) placed in a rarer medium of refractive index \(n_1\) in front of a concave refracting spherical surface of radius of curvature \(R\). A virtual, erect image \(I\) is formed at distance \(v\) in the same medium.
Let \(A\) be the point of incidence, and \(NC\) be the normal to the surface at \(A\).
Let \(\alpha, \beta, \gamma\) be the angles made by incident ray, refracted ray, and normal with the principal axis.
From triangle \(AOC\), exterior angle \(i = \gamma - \alpha\).
From triangle \(AIC\), exterior angle \(r = \gamma - \beta\).
According to Snell's law: 
\(n_1 \sin i = n_2 \sin r\)
For small apertures, \(\sin i \approx i\) and \(\sin r \approx r\):
\(n_1 i = n_2 r\)
Substitute the values of \(i\) and \(r\):
\(n_1(\gamma - \alpha) = n_2(\gamma - \beta)\)
\(n_2\beta - n_1\alpha = (n_2 - n_1)\gamma\)
Using small angle approximations (in radians, \(\tan \theta \approx \theta\)):
\(\alpha \approx \frac{AM}{-u}, \quad \beta \approx \frac{AM}{-v}, \quad \gamma \approx \frac{AM}{-R}\)
Substituting these into the equation and cancelling \(AM\):
\(n_2\left(\frac{-1}{v}\right) - n_1\left(\frac{-1}{u}\right) = (n_2 - n_1)\left(\frac{-1}{R}\right)\)
Multiplying throughout by \(-1\):
\(\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}\)
In simple words: By applying Snell's law, exterior angle geometry in triangles, and small-angle approximations, we successfully derive the refraction formula for spherical surfaces.

Exam Tip: Always follow sign conventions strictly (distances measured against incident light are negative).

Question 12. Derive the relation \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\) for a convex lens. Draw \(m\) vs \(v\) graph. Write how we can find focal length from this graph?
Answer: Consider a thin convex lens of focal length \(f\). An object \(AB\) is placed beyond focus at distance \(u\), forming a real, inverted image \(A'B'\) at distance \(v\).
From similar triangles \(\triangle ABC\) and \(\triangle A'B'C\) (where \(C\) is optical center):
\(\frac{A'B'}{AB} = \frac{CB'}{CB} = \frac{v}{-u}\)
From similar triangles \(\triangle CDF\) and \(\triangle A'B'F\):
\(\frac{A'B'}{CD} = \frac{FB'}{CF} \implies \frac{A'B'}{AB} = \frac{v - f}{f}\)
Equating both expressions for magnification:
\(\frac{-v}{u} = \frac{v - f}{f}\)
\(-vf = uv - uf\)
Dividing the entire equation by \(uvf\):
\(-\frac{1}{u} = \frac{1}{f} - \frac{1}{v} \implies \frac{1}{f} = \frac{1}{v} - \frac{1}{u}\)

Magnification (\(m\)) vs Image distance (\(v\)) graph:
For a convex lens, magnification is given by \(m = \frac{v - f}{f} = 1 - \frac{v}{f}\) or \(m = \frac{f}{f + u}\).
When plotting \(m\) versus \(v\):
- The graph is a straight line with a negative slope equal to \(-\frac{1}{f}\).
- At \(v = f\), \(m = 0\).
- The intercept on the \(m\) axis is \(1\), and the intercept on the \(v\) axis gives the focal length \(f\).
In simple words: Using similar triangles for a convex lens gives the lens formula. The \(m\) vs \(v\) graph is a straight line where the intercepts help directly calculate the focal length.

Exam Tip: Label the axes, intercepts, and slope clearly when sketching graphical questions in optics.

Question 13. Derive an expression for angle of deviation for an equilateral prism. Draw a graph between angle of deviation vs angle of incidence. Write the conditions of minimum deviation. Derive a relation for refractive index of the material of prism.
Answer: Derivation of angle of deviation:\n Let a ray of light incident at angle \(i_1\) on face \(AB\) of a prism refracts at angle \(r_1\), strikes the second face at angle of incidence \(r_2\), and emerges at angle \(i_2\).
The total deviation \(\delta\) produced by the prism is the sum of deviations at both refracting surfaces:
\(\delta = (i_1 - r_1) + (i_2 - r_2) = (i_1 + i_2) - (r_1 + r_2)\)
From the quadrilateral geometry of the prism, \(r_1 + r_2 = A\) (where \(A\) is the prism angle).
Therefore, \(\delta = i_1 + i_2 - A\).

Conditions for minimum deviation (\(\delta_m\)):
1. The angle of incidence equals the angle of emergence (\(i_1 = i_2 = i\)).
2. The angle of refraction inside the prism are equal (\(r_1 = r_2 = r = A/2\)).
3. The refracted ray inside the prism passes parallel to the base of the prism.

Refractive Index Derivation:
At minimum deviation: \(\delta = \delta_m\), \(i_1 = i = \frac{A + \delta_m}{2}\) and \(r_1 = r = \frac{A}{2}\).
Using Snell's law, the refractive index of the prism material is:
\(\mu = \frac{\sin i}{\sin r} = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}\)
In simple words: Deviation is the sum of entry and exit deviations minus the prism angle. At minimum deviation, symmetric bending occurs, leading to the standard prism refractive index formula.

Exam Tip: Practice the \(\delta\) vs \(i\) asymmetric curve; remember it drops to a minimum \(\delta_m\) before rising again.

Question 14. Draw the working of compound microscope. Derive the relation for its magnifying power. Why we use the lens of small aperture as objective in compound microscope?
Answer: Magnifying Power Derivation:
Magnifying power \(m\) is defined as the ratio of the angle subtended by the final image at the eye to the angle subtended by the object seen directly at the least distance of distinct vision (\(D\)):
\(m = \frac{\beta}{\alpha} \approx \frac{\tan\beta}{\tan\alpha}\)
From the geometry of the compound microscope lenses:
Magnification can be expressed as the product of magnification produced by the objective lens (\(m_o\)) and the eyepiece (\(m_e\)):
\(m = m_o \times m_e = \left(\frac{h'}{h}\right) \times \left(1 + \frac{D}{f_e}\right) = \frac{-v_o}{u_o}\left(1 + \frac{D}{f_e}\right)\)
When the final image is formed at infinity (normal adjustment):
\(m = \frac{-v_o}{u_o} \cdot \frac{D}{f_e}\)

Why use a lens of small aperture as objective?
A small aperture objective lens reduces spherical and chromatic aberrations, ensuring a sharper, more focused image, and collects light from fine microscopic details effectively.
In simple words: A compound microscope uses two lenses to multiply magnification. A small aperture objective is used to avoid optical distortions and ensure high clarity.

Exam Tip: Clearly state whether the final image is formed at the least distance of distinct vision (\(D\)) or at infinity when writing final magnification formulas.

Question 15. (A) Why we use lens of greater focal length as objective and lens of small focal length as an eyelens in compound microscope?
Answer: Note: In a compound microscope, the objective actually has a very *small* focal length (and small aperture) to produce a magnified real image, while the eyepiece has a moderate focal length. (If referring to a telescope: objective has large focal length to gather maximum light from distant stars).
In simple words: Objectives have small focal lengths in microscopes to provide high primary magnification of tiny nearby objects.

Exam Tip: Double-check whether the instrument is a microscope (small objective focal length) or a telescope (large objective focal length).

Question 15. (B) Why we use lenses of small focal lengths in compound microscope?
Answer: Lenses of small focal lengths have higher optical power, which allows both the objective and eyepiece to produce greater linear and angular magnification, enabling the visualization of extremely small microscopic specimens.
In simple words: Smaller focal length means higher power, which multiplies magnification to view microscopic objects clearly.

Exam Tip: High power equals short focal length (\(P = 1/f\)).

Question 16. (A) For which position of an object in front of convex lens of power +10D it will form an image which is twice in size of object.
Answer: Given power of convex lens \(P = +10\text{ D}\).
Focal length \(f = \frac{1}{P} = \frac{1}{10}\text{ m} = +10\text{ cm}\).
Magnification \(m = \frac{v}{u}\). 
For a convex lens forming a real image, magnification is \(-2\):
\(-2 = \frac{v}{u} \implies v = -2u\)
Using lens formula \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\):
\(\frac{1}{10} = \frac{1}{-2u} - \frac{1}{u} = \frac{-1 - 2}{2u} = \frac{-3}{2u}\)
\(2u = -30 \implies u = -15\text{ cm}\)
For a virtual image, magnification is \(+2\):
\(+2 = \frac{v}{u} \implies v = 2u\)
\(\frac{1}{10} = \frac{1}{2u} - \frac{1}{u} = \frac{1 - 2}{2u} = \frac{-1}{2u}\)
\(2u = -10 \implies u = -5\text{ cm}\)
Thus, the object can be placed either at \(15\text{ cm}\) (for real image) or \(5\text{ cm}\) (for virtual image) in front of the lens.
In simple words: By testing both real (\(m = -2\)) and virtual (\(m = +2\)) image conditions with the lens formula, we find the object must be placed at \(15\text{ cm}\) or \(5\text{ cm}\).

Exam Tip: Always consider both real and virtual image cases when magnification magnitude is given without sign.

Question 16. (B) Write the differences between interference and diffraction.
Answer:
1. Origin: Interference is the superposition of two separate wavefronts originating from two coherent sources. Diffraction is the superposition of secondary wavelets from different parts of the same wavefront.
2. Fringe Width: In interference, all interference fringes are of equal width. In diffraction, fringes are of varying width with a prominent central maximum.
3. Intensity: In interference, all bright fringes have uniform intensity. In diffraction, the intensity falls rapidly as we move away from the central maximum.
In simple words: Interference is between two separate waves with equal fringes, while diffraction is bending around corners resulting in a wide central bright fringe with decreasing brightness.

Exam Tip: Memorize at least 3 distinct comparison points (source, fringe width, intensity variation) for interference vs diffraction.