CBSE Class 12 Physics Wave Optics MCQs Set 07

Answer: (b) spherical

Explanation:
1. A point source sends out disturbances equally in all directions.
2. All points at the same distance from the source are in the same phase, so the wavefronts are spherical.

Teacher's Note:
1. A slit (linear source) gives cylindrical wavefronts.
2. Far from any source, a small part of the wavefront looks plane.

Answer: (c) 400 nm and 5 × 10¹⁴ Hz

Explanation:
1. Frequency ν = \( \frac{c}{\lambda} \) = \( \frac{3 \times 10^8}{600 \times 10^{-9}} \) = 5 × 10¹⁴ Hz, and it does not change on refraction.
2. λ_glass = \( \frac{\lambda}{n} \) = \( \frac{600}{1.5} \) = 400 nm
3. Answer: 400 nm and 5 × 10¹⁴ Hz

Teacher's Note:
1. Frequency belongs to the source; speed and wavelength change with the medium.
2. Multiplying by n instead of dividing gives the trap answer 900 nm.

Answer: (a) 22°

Explanation:
1. \( \frac{\sin i}{\sin r} = \frac{v_1}{v_2} \) ⇒ sin r = \( \frac{v_2}{v_1} \) sin i = \( \frac{1.5}{2} \) × 0.5 = 0.375
2. r = sin⁻¹(0.375) ≈ 22°

Teacher's Note:
1. Light slows down in medium 2, so it bends towards the normal (r < i).
2. Put the speeds in the right order: sin i/sin r = v₁/v₂.

Answer: (d) the forward envelope of the secondary wavelets

Explanation:
1. Every point on the old wavefront sends out secondary wavelets of radius vτ.
2. The common tangent drawn to these wavelets in the forward direction is the new wavefront.
3. Answer: the forward envelope of the secondary wavelets

Teacher's Note:
1. Huygens ignored the backward wave by assuming the wavelets have zero amplitude backwards.
2. The wavelets spread with the speed of light in that medium, not always c.

Answer: (b) 1 mm

Explanation:
1. β = \( \frac{\lambda D}{d} \) = \( \frac{500 \times 10^{-9} \times 1}{0.5 \times 10^{-3}} \)
2. β = 1 × 10⁻³ m = 1 mm

Teacher's Note:
1. Convert nm and mm to metres before substituting.
2. Fringe widths in YDSE usually come out as a fraction of a millimetre to a few millimetres.

Answer: (d) 3.0 mm

Explanation:
1. Dark fringes lie at x = \( \left(n + \frac{1}{2}\right)\beta \) with n = 0, 1, 2, …; the third dark fringe has n = 2.
2. x = 2.5 × 1.2 = 3.0 mm

Teacher's Note:
1. The first dark fringe is at β/2, not at β.
2. 3.6 mm is the position of the third bright fringe.

Answer: (a) 0.45 mm

Explanation:
1. In water the wavelength becomes λ/n, so β′ = \( \frac{\beta}{n} \).
2. β′ = \( \frac{0.6}{4/3} \) = 0.6 × 0.75 = 0.45 mm

Teacher's Note:
1. The fringes come closer together in a denser medium.
2. D and d are unchanged; only λ changes.

Answer: (c) 4 : 1

Explanation:
1. Amplitude ratio a₁ : a₂ = \( \sqrt{9} : \sqrt{1} \) = 3 : 1
2. \( \frac{I_{\max}}{I_{\min}} = \frac{(3 + 1)^2}{(3 - 1)^2} = \frac{16}{4} \) = 4 : 1

Teacher's Note:
1. Always take square roots of intensities first to get the amplitude ratio.
2. Equal intensities would give a perfectly dark minimum (ratio infinite).

Answer: (b) I₀

Explanation:
1. Phase difference φ = \( \frac{2\pi}{\lambda} \times \frac{\lambda}{3} \) = \( \frac{2\pi}{3} \)
2. I = 4I₀ cos²\( \left(\frac{\pi}{3}\right) \) = 4I₀ × \( \frac{1}{4} \) = I₀

Teacher's Note:
1. Convert path difference to phase difference before using the intensity formula.
2. Remember to halve φ inside the cos² term.

Answer: (a) the phase difference between their waves changes rapidly and randomly

Explanation:
1. For a steady pattern the two sources must keep a constant phase difference (coherent sources).
2. Light from independent lamps is emitted by different atoms at random instants, so the phase difference changes about 10⁸ times a second.
3. The pattern shifts too fast to be seen and only uniform illumination appears. Answer: the phase difference changes rapidly and randomly.

Teacher's Note:
1. Both lamps give the same yellow light, so 'different colours' is wrong.
2. Young solved this by deriving two sources from one slit.

Answer: (d) becomes wider

Explanation:
1. The narrower the aperture, the more the light spreads out on passing through it.
2. So the central maximum becomes wider (and less bright).

Teacher's Note:
1. Diffraction is strongest when the slit width is comparable to the wavelength.
2. Treat this only qualitatively; no single-slit formula is needed.

Answer: (c) two narrow slits illuminated by the same monochromatic source

Explanation:
1. Equally spaced maxima of equal height are the signature of interference between two coherent sources.
2. A single slit gives a broad central maximum with much weaker side maxima, and independent bulbs give uniform illumination.
3. Answer: two narrow slits illuminated by the same monochromatic source (Young's experiment).

Teacher's Note:
1. Compare the shape with the single-slit pattern: there the central band is twice as wide.
2. A polaroid only changes intensity; it does not make fringes.

Answer: (b) I₀/8

Explanation:
1. After the first polaroid: I₁ = \( \frac{I_0}{2} \)
2. By Malus' law: I = I₁ cos² 60° = \( \frac{I_0}{2} \times \frac{1}{4} \) = \( \frac{I_0}{8} \)

Teacher's Note:
1. Malus' law applies only to light that is already polarised.
2. Forgetting the first halving gives the wrong answer I₀/4.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. On refraction the speed and wavelength change but the frequency stays the same, so the Assertion is true.
2. The number of oscillations per second is fixed by the source, so the Reason is true.
3. The Reason correctly explains the Assertion.

Teacher's Note:
1. v = νλ: if v falls, λ falls in the same ratio.
2. This is why the colour of light does not change in glass.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. n_glass ≈ 1.5, so v = c/n ≈ 2 × 10⁸ m s⁻¹, less than in air. The Assertion is true.
2. Huygens' construction does explain both laws, so the Reason is true.
3. The Reason is a general statement about the theory; it does not explain why light is slower in glass.

Teacher's Note:
1. The wave theory's prediction of slower light in water was confirmed by experiment.
2. Choose (b) when the Reason is true but not the cause.

Answer: (c) A is true, but R is false.

Explanation:
1. The centre is equidistant from both slits, so the waves arrive in phase and the central fringe is bright. The Assertion is true.
2. The path difference at the centre is zero, not λ/2. A path difference of λ/2 would give a dark fringe. The Reason is false.

Teacher's Note:
1. Zero path difference holds for all wavelengths, which is why the central fringe is white in white light.
2. Path difference λ/2 always means destructive interference.

Answer: (d) A is false, but R is true.

Explanation:
1. Sound waves (wavelength about 1 m) bend round doors and walls, but light rarely shows noticeable bending. The Assertion is false.
2. Light has a wavelength of about 5 × 10⁻⁷ m, much smaller than everyday objects. The Reason is true.

Teacher's Note:
1. Diffraction is noticeable only when the opening is comparable to the wavelength.
2. The true Reason actually explains why the Assertion is false.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The transmitted intensity of unpolarised light through one polaroid is I₀/2, so the Assertion is true.
2. Unpolarised light has vibrations in all directions; the polaroid passes I₀cos²θ for each, and the average of cos²θ is ½. The Reason is true.
3. The Reason explains the Assertion.

Teacher's Note:
1. This result is independent of how the polaroid is rotated.
2. The light that comes out is plane polarised.

Answer: (c) A is true, but R is false.

Explanation:
1. β = λD/d and λ becomes λ/n in water, so β decreases. The Assertion is true.
2. The frequency of light does not change on entering water; only its speed and wavelength decrease. The Reason is false.

Teacher's Note:
1. β′ = β/n for an apparatus immersed in a medium.
2. Frequency never changes on refraction.

Answer: (a) 3.6 mm

Explanation:
1. From the 1st to the 11th bright fringe there are 10 fringe widths.
2. β = \( \frac{36}{10} \) = 3.6 mm

Teacher's Note:
1. Count the gaps, not the fringes: 11 − 1 = 10.
2. Dividing by 11 gives the trap answer 3.3 mm.

Answer: (c) 600 nm

Explanation:
1. λ = \( \frac{\beta d}{D} \) = \( \frac{3.6 \times 10^{-3} \times 0.25 \times 10^{-3}}{1.5} \)
2. λ = 6 × 10⁻⁷ m = 600 nm

Teacher's Note:
1. Rearrange β = λD/d to make λ the subject.
2. 600 nm is in the orange-red part of the spectrum, as expected for a red laser.

Answer: (b) 4.8 mm

Explanation:
1. β ∝ D, so β′ = 3.6 × \( \frac{2.0}{1.5} \)
2. β′ = 4.8 mm

Teacher's Note:
1. Moving the screen away spreads the fringes out.
2. Use ratios to save time when only one quantity changes.

Answer: (d) 5.4 mm

Explanation:
1. Dark fringes are at x = \( \left(n + \frac{1}{2}\right)\beta \); the second dark fringe has n = 1.
2. x = 1.5 × 3.6 = 5.4 mm

Teacher's Note:
1. The first dark fringe is at 1.8 mm, half a fringe width from the centre.
2. 7.2 mm is the second bright fringe.

Answer: (a) The interference fringes disappear and a broad diffraction band of the single slit is seen

Explanation:
1. Interference needs light from two coherent sources; with one slit covered there is nothing to interfere with.
2. Light from the single open slit still spreads by diffraction, so a broad patch of light is seen.
3. Answer: the interference fringes disappear and a broad single-slit diffraction band is seen.

Teacher's Note:
1. The total light reaching the wall is also reduced to about half.
2. Interference fringes are seen only when both slits are open.

Answer: (b) 2.25 × 10⁸ m s⁻¹

Explanation:
1. v = \( \frac{c}{n} \) = \( \frac{3 \times 10^8}{4/3} \)
2. v = 2.25 × 10⁸ m s⁻¹

Teacher's Note:
1. No light can travel faster than c, so 4 × 10⁸ m s⁻¹ is impossible.
2. Dividing by 4/3 is the same as multiplying by 0.75.

Answer: (d) 473 nm

Explanation:
1. λ_water = \( \frac{\lambda}{n} \) = \( \frac{630}{4/3} \) = 630 × 0.75
2. λ_water ≈ 473 nm

Teacher's Note:
1. The wavelength shrinks in the same ratio as the speed.
2. The light still looks red to an eye inside the water, because colour depends on frequency.

Answer: (a) 37°

Explanation:
1. sin r = \( \frac{\sin i}{n} \) = \( \frac{0.8}{4/3} \) = 0.6
2. r = sin⁻¹(0.6) ≈ 37°

Teacher's Note:
1. Going into a denser medium, the ray bends towards the normal, so r must be less than 53°.
2. sin 37° ≈ 0.6 and sin 53° ≈ 0.8 are useful values to remember.

Answer: (c) 4.76 × 10¹⁴ Hz

Explanation:
1. ν = \( \frac{c}{\lambda} \) = \( \frac{3 \times 10^8}{630 \times 10^{-9}} \) ≈ 4.76 × 10¹⁴ Hz in air.
2. The frequency is unchanged in water, so ν ≈ 4.76 × 10¹⁴ Hz.

Teacher's Note:
1. Check: v/λ in water gives the same value, 2.25 × 10⁸ / 473 × 10⁻⁹.
2. Do not convert nm with 10⁻⁶; 1 nm = 10⁻⁹ m.

Answer: (b) the same for all rays

Explanation:
1. A wavefront joins points in the same phase, so light must take equal time from one wavefront to the next along every ray.
2. A ray with a longer path in air has a shorter path in water, and the times balance.
3. Answer: the same for all rays

Teacher's Note:
1. This equal-time idea is how Huygens' construction gives Snell's law.
2. The same idea explains how a lens brings all rays to one focus.