Physics · Optics · chapter 10 of 14

Class 12 Physics Chapter 10: Wave Optics worksheet with answer key

A 20-mark worksheet on Wave Optics for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.

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Question 1
A beam of light passes from glass into air. The quantity that does not change is its
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Answer: (a) frequency

In the Huygens construction a crest from A reaches E as the crest from B reaches C, so v₁/λ₁ = v₂/λ₂: the frequency is unchanged, while speed and wavelength both increase in air and an oblique ray bends.

Question 2
Light of frequency 6.0 × 10¹⁴ Hz has a wavelength of 400 nm inside a transparent plastic. Taking c = 3.0 × 10⁸ m s⁻¹, the refractive index of the plastic is
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Answer: (c) 1.25

The frequency is the same in every medium, so the speed in the plastic is v = νλ = 6.0 × 10¹⁴ × 400 × 10⁻⁹ = 2.4 × 10⁸ m s⁻¹, and n = c/v = 3.0/2.4 = 1.25. 0.80 is v/c, inverted; 1.33 is the index of water, assumed rather than worked out; 2.40 is the speed in units of 10⁸ m s⁻¹.

Question 3
According to the superposition principle, the resultant displacement at a point where several waves meet is
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Answer: (a) the vector sum of the displacements due to each wave

The resultant displacement is the vector sum of the displacements produced by each wave. Intensities add only for incoherent sources, and that is a time average, not a displacement.

Question 4
A convex lens forms a real image of a point object. The ray through the centre of the lens covers a shorter geometrical path than a ray through the edge, yet both reach the image together because
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Answer: (d) the central ray spends more of its path in glass, where light travels more slowly

The time from object point to image point is the same along every ray. The central ray's path is shorter, but more of it lies in glass, where light is slower; the two effects balance.

Question 5
Light travels at 2.0 × 10⁸ m s⁻¹ in a certain glass. Taking c = 3.0 × 10⁸ m s⁻¹, the critical angle for light going from this glass into air is
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Answer: (b) 41.8°

n = c/v = 1.5, and sin ic = n₂/n₁ = 1/1.5 = 2/3, so ic ≈ 41.8°. 48.2° is the complement; 33.7° takes tan ic = 2/3; 19.5° takes sin ic = 1/3.

Question 6 · 2 marks
Two coherent sources each produce an intensity I₀ at a point on a screen. Find the resultant intensity at points where the phase difference between the two waves is (i) π/2, (ii) 5π/3.
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I = 4I₀ cos²(φ/2). (i) φ = π/2: I = 4I₀ cos²(π/4) = 4I₀ × 1/2 = 2I₀. (ii) φ = 5π/3: I = 4I₀ cos²(5π/6) = 4I₀ × 3/4 = 3I₀.

  • I = 4I₀cos²(φ/2); (i) 2I₀ (1)
  • (ii) 3I₀ (1)
Question 7 · 2 marks
State Huygens' principle. Using it, describe how the position of a plane wavefront at a time τ later is obtained, and state the direction of the rays.
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Every point of a wavefront is a source of secondary wavelets, which spread out in all directions with the speed of the wave; the common tangent (forward envelope) of these wavelets at a later time gives the new wavefront. For a plane wavefront F₁F₂, draw spheres of radius vτ about several of its points; the plane G₁G₂ touching them is the wavefront at time τ, parallel to F₁F₂. The rays are the lines normal to both wavefronts.

  • each point a source of secondary wavelets moving at the wave speed; their envelope is the new wavefront (1)
  • spheres of radius vτ give the parallel plane G₁G₂; rays are normal to the wavefronts (1)
Question 8 · 2 marks
Using the Huygens construction for the refraction of a plane wavefront at a plane surface, show that λ₁/λ₂ = v₁/v₂, and hence that the frequency of the light does not change. State how the wavelength changes when light passes into a denser medium.
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In the construction, while the edge B of the incident wavefront AB travels BC = v₁τ to the surface at C, the wavelet from A spreads to radius AE = v₂τ in the second medium; CE is the refracted wavefront. Take τ as one period: then BC is one wavelength λ₁ and AE one wavelength λ₂, since the crest from A reaches E as the crest from B reaches C. So λ₁/λ₂ = v₁/v₂, that is v₁/λ₁ = v₂/λ₂: the frequency v/λ is the same in both media. In a denser medium v₂ < v₁, so λ₂ < λ₁: the wavelength becomes shorter in the same ratio as the speed.

  • BC = v₁τ, AE = v₂τ; with τ one period, BC = λ₁ and AE = λ₂, so λ₁/λ₂ = v₁/v₂ and v₁/λ₁ = v₂/λ₂, frequency unchanged (1)
  • denser medium: v₂ < v₁, so the wavelength decreases in proportion to the speed (1)
Question 9 · 3 marks
In a double-slit experiment the slits are 0.35 mm apart, the screen is 1.25 m away and the light has a wavelength of 560 nm. Find (i) the fringe width, (ii) the distance between the second bright fringe on one side of the centre and the third dark fringe on the other side, (iii) the fringe width when the screen is moved to 1.75 m.
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(i) β = λD/d = 560 × 10⁻⁹ × 1.25/(0.35 × 10⁻³) = 2.0 × 10⁻³ m = 2.0 mm. (ii) Second bright fringe at 2β = 4.0 mm on one side; third dark fringe at (2 + 1/2)β = 5.0 mm on the other; separation 9.0 mm. (iii) β ∝ D: β = 2.0 × 1.75/1.25 = 2.8 mm.

  • β = λD/d = 2.0 mm (1)
  • 2β + 2.5β = 9.0 mm (1)
  • β ∝ D gives 2.8 mm (1)
Question 10 · 3 marks
Unpolarised light of intensity 32 W m⁻² falls on a polaroid P₁, and the light from it on a second polaroid P₂. The light emerging from P₂ has intensity 4.0 W m⁻². Find (i) the intensity between P₁ and P₂, (ii) the angle between their pass-axes, (iii) the smallest angle through which P₂ must be turned to double the emerging intensity.
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(i) P₁ halves unpolarised light: 16 W m⁻². (ii) Malus' law: 4.0 = 16 cos²θ, cos²θ = 1/4, θ = 60°. (iii) For 8.0 W m⁻², cos²θ′ = 1/2, θ′ = 45°: turn P₂ through 15°, bringing its pass-axis closer to that of P₁.

  • 16 W m⁻² after P₁ (1)
  • cos²θ = 1/4, θ = 60° (1)
  • cos²θ′ = 1/2, θ′ = 45°, turn by 15° (1)
Question 11 · 3 marks
Light travels at 2.25 × 10⁸ m s⁻¹ in a liquid. It meets the plane surface between the liquid and air from inside the liquid. Taking c = 3.00 × 10⁸ m s⁻¹, find (i) the refractive index of the liquid, (ii) the critical angle for the liquid–air surface. (iii) Using the Huygens construction, explain why there is no refracted wavefront when the angle of incidence exceeds the critical angle.
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(i) n = c/v = 3.00/2.25 = 1.33 (4/3). (ii) sin ic = n₂/n₁ = 1/(4/3) = 0.75, ic ≈ 48.6°. (iii) Going into the rarer medium, v₂ > v₁, so the wavelet from A in air has radius v₂τ greater than BC = v₁τ. The refracted wavefront is the tangent from C to this wavelet, with sin r = (v₂/v₁) sin i. When sin i > v₁/v₂, the wavelet's radius v₂τ exceeds AC, so C lies inside it and no tangent plane can be drawn from C (sin r would exceed 1): there is no refracted wave, and the light is totally reflected.

  • n = c/v = 4/3 (1)
  • sin ic = 3/4, ic ≈ 48.6° (1)
  • for i > ic the wavelet radius v₂τ exceeds AC; no tangent from C, so no refracted wavefront (1)
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