Chapter 10: Wave Optics Application Questions | physics Class 12 CBSE

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Chapter 10: Wave Optics Application Questions | physics Class 12 CBSE | ByteNotes.in

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 1

Applied Concept

A submarine uses optical fibres for internal communication. With reference to the wave model of light, explain the physical principle that allows light to travel through a bent optical fibre without escaping into the surrounding medium.

Optical fibres exploit total internal reflection: light travels in the denser glass core and strikes the glass-cladding interface (cladding has lower refractive index) at angles exceeding the critical angle ic = sin⁻¹(n2/n1), so no light is refracted out — it is completely reflected back into the core.
As long as the bends in the fibre are gentle enough that the angle of incidence at every point on the core-cladding interface remains above ic, the light undergoes repeated total internal reflections and propagates along the fibre regardless of its shape.
The wave model explains why frequency is preserved along the fibre (boundary conditions maintain frequency), while speed and wavelength adjust to the refractive index of the core, ensuring efficient, low-loss signal transmission over long distances.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 2

Applied Concept

In a Young's double-slit experiment conducted in a school lab, a student observes that when the entire apparatus is immersed in water (refractive index 1.33), the fringe width decreases. Using the formula for fringe width, quantitatively explain this observation.

The fringe width in any medium is β = λ_medium × D/d, where λ_medium = λ_air/n. When immersed in water (n = 1.33), the wavelength decreases to λ_water = λ_air/1.33, so β_water = λ_air D / (1.33 d) = β_air / 1.33.
The fringe width in water is therefore 1/1.33 ≈ 0.75 times the fringe width in air — the fringes become about 25% narrower. This occurs because the speed of light in water is lower, reducing the wavelength while frequency remains unchanged.
This result confirms that fringe width depends on the wavelength of light in the medium of propagation, not the vacuum wavelength; the student can use the measured reduction in fringe width to verify the refractive index of the water.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 3

Applied Concept

A photographer wants to take a picture of a lake without glare from the reflected sunlight. Explain, with reference to the polarisation of reflected light, how a polaroid filter attached to the camera lens helps achieve this.

Sunlight reflected from a horizontal water surface at shallow angles is predominantly horizontally polarised — the reflected electric field has a strong horizontal component. This polarised glare overwhelms the actual scene beneath the water surface.
A polaroid filter mounted on the camera lens with its pass-axis oriented vertically blocks the horizontally polarised reflected light while transmitting the (unpolarised or differently polarised) light coming from the scene, effectively eliminating the glare.
By rotating the polaroid filter until reflected glare is minimised, the photographer can observe and photograph the lake bed or underwater objects that would otherwise be invisible; the intensity of glare follows Malus' law (I = I0 cos²θ) as the filter is rotated.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 4

Applied Concept

With reference to Huygens principle, explain why a sound wave (not light) can be heard around a corner of a building, while a light source around the same corner is not visible.

Diffraction — bending of waves around obstacles — is significant when the wavelength is comparable to or larger than the obstacle dimensions. Sound waves have wavelengths of 1 cm to several metres, which are comparable to the size of buildings and doorways, so sound diffracts appreciably around corners.
Visible light has wavelengths of only 400–700 nm, which are millions of times smaller than the corner of a building; the diffraction angle θ ~ λ/a is vanishingly small, so light does not appreciably bend around the corner and the corner casts a sharp geometric shadow.
Huygens principle predicts diffraction for both: every point on the wavefront at the corner acts as a secondary source, but for light the secondary wavelets are so tightly clustered in the forward direction (λ/a → 0) that the side spread is imperceptible, while for sound the spread is large.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 5

Applied Concept

A student performing a single-slit diffraction experiment observes that the central bright fringe becomes wider when the slit width is decreased. Using the diffraction formula, explain this observation and state the limiting case.

The angular half-width of the central maximum in single-slit diffraction is θ1 = λ/a; the linear width on a screen at distance D is 2λD/a. When the slit width a is decreased, the central maximum width 2λD/a increases — the pattern spreads out more.
This is a direct consequence of the wave nature of light: a narrower slit means fewer secondary sources spanning a shorter width, and by Huygens principle these sources create secondary wavelets that spread more widely in the lateral direction.
In the limiting case as a → 0, the central maximum spreads over the entire screen and the slit acts as a nearly isotropic point source; as a → ∞ (very wide slit), the central maximum narrows to a very sharp beam and geometrical optics (no diffraction) is recovered.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 6

Applied Concept

The rainbow colours seen on the surface of a CD when it is held under white light are explained by wave optics. Explain the physical mechanism responsible, with reference to the track spacing on the CD.

A CD surface has closely spaced parallel tracks (spacing ~ 1.6 μm) that act as a reflection diffraction grating. When white light (containing all visible wavelengths, 400–700 nm) falls on the surface, each wavelength is diffracted at a different angle according to the grating equation, separating the colours spatially.
The track spacing (~1.6 μm) is comparable to the wavelengths of visible light, satisfying the condition for significant diffraction. Different colours (wavelengths) reach the eye from different angles as the disc is tilted, producing the characteristic rainbow pattern.
This is a diffraction phenomenon, not refraction: no medium boundary is crossed. The colours change with viewing angle because the diffraction angle θ depends on wavelength (longer λ → larger θ), so different colours are bright at different tilt angles.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 7

Applied Concept

In a Young's double-slit experiment, if one slit is covered with a red filter and the other with a blue filter, will interference fringes be observed? Justify using the concept of coherence and the condition for interference.

No stable interference fringes will be observed. A red filter transmits only red light (λ ~ 650 nm) and a blue filter transmits only blue light (λ ~ 470 nm); the two slits now emit light of different frequencies (different wavelengths).
For a stable interference pattern, the two sources must be coherent — they must have the same frequency and a constant phase difference. Sources emitting different frequencies produce an interference term that oscillates in time at the beat frequency, averaging to zero over the detector response time.
Since red and blue light have different frequencies, the phase difference between them at any point on the screen varies rapidly with time (at the frequency difference of ~1.4 × 10¹⁴ Hz); the time-averaged intensity is uniform at I1 + I2, with no fringe pattern visible.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 8

Applied Concept

A 3D cinema uses two projectors with polaroids whose pass-axes are perpendicular to each other, and the audience wears polaroid glasses. Explain the principle by which each eye sees only one of the two projected images.

The two projectors transmit two slightly different perspective images of the same scene, each through a polaroid: projector 1 uses a polaroid with pass-axis horizontal (transmitting horizontally polarised light) and projector 2 uses a vertical pass-axis (transmitting vertically polarised light).
The audience wears glasses where the left lens has a horizontal pass-axis polaroid and the right lens has a vertical pass-axis polaroid (or vice versa). Each lens transmits only the light from the matching projector and blocks the other (crossed polaroids transmit zero intensity per Malus' law at θ = 90°).
Thus the left eye sees only the left-perspective image and the right eye sees only the right-perspective image, providing the binocular disparity that the brain interprets as three-dimensional depth — the entire effect relying on the polarisation of light.

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Application Question

Easy difficulty • Concept in a practical situation

Easy

Question 9

Applied Concept

With reference to the experiment described in the chapter, explain how a student can observe a single-slit diffraction pattern at home using two razor blades and a bulb with a straight filament.

The student holds two razor blades with edges parallel, creating a narrow adjustable slit between them. The straight filament of a clear glass bulb acts as a coherent line source (analogous to the first slit S in the formal experiment); the slit formed by the blades acts as the single diffracting aperture.
Light from the filament passes through the narrow blade slit and diffracts; the lens of the eye focuses the diffracted light onto the retina (acting as the screen). With slight adjustment of slit width and parallelism, the central bright band flanked by alternate dark and bright regions is observed.
Since the positions of bands (except central) depend on wavelength, the pattern shows some colour; using a red or blue filter isolates one wavelength and makes the fringes cleaner. Red light produces wider fringes than blue (β ∝ λ), demonstrating the wavelength dependence of diffraction.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 10

Applied Concept

A beam of monochromatic light passes from air into a glass slab and then back into air. Using the wave model, describe what happens to the wavelength, speed, and frequency of light at each interface, and explain why the ray emerges parallel to the incident ray.

At the air-glass interface: frequency remains unchanged (boundary condition), but speed decreases from c to c/n and wavelength decreases from λ to λ/n (where n = refractive index of glass). Snell's law n1 sin i = n2 sin r applies; the ray bends toward the normal.
Inside the glass: the light travels at reduced speed v = c/n with wavelength λ_glass = λ/n, still at the same frequency ν. The wave maintains its character as it propagates through the uniform glass.
At the glass-air interface: speed and wavelength increase back to their original air values (c and λ), and by Snell's law the ray bends away from the normal by the same angle as it bent toward the normal entering — the two refractions are equal and opposite. Since the two faces of the slab are parallel, the emerging ray is parallel to the incident ray but laterally displaced.

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Chapter 10: Wave Optics | Application Questions

A submarine uses optical fibres for internal communication. With reference to the wave model of light, explain the physical principle that allows light to travel through a bent optical fibre without escaping into the surrounding medium.

In a Young's double-slit experiment conducted in a school lab, a student observes that when the entire apparatus is immersed in water (refractive index 1.33), the fringe width decreases. Using the formula for fringe width, quantitatively explain this observation.

A photographer wants to take a picture of a lake without glare from the reflected sunlight. Explain, with reference to the polarisation of reflected light, how a polaroid filter attached to the camera lens helps achieve this.

With reference to Huygens principle, explain why a sound wave (not light) can be heard around a corner of a building, while a light source around the same corner is not visible.

A student performing a single-slit diffraction experiment observes that the central bright fringe becomes wider when the slit width is decreased. Using the diffraction formula, explain this observation and state the limiting case.

The rainbow colours seen on the surface of a CD when it is held under white light are explained by wave optics. Explain the physical mechanism responsible, with reference to the track spacing on the CD.

In a Young's double-slit experiment, if one slit is covered with a red filter and the other with a blue filter, will interference fringes be observed? Justify using the concept of coherence and the condition for interference.

A 3D cinema uses two projectors with polaroids whose pass-axes are perpendicular to each other, and the audience wears polaroid glasses. Explain the principle by which each eye sees only one of the two projected images.

With reference to the experiment described in the chapter, explain how a student can observe a single-slit diffraction pattern at home using two razor blades and a bulb with a straight filament.

A beam of monochromatic light passes from air into a glass slab and then back into air. Using the wave model, describe what happens to the wavelength, speed, and frequency of light at each interface, and explain why the ray emerges parallel to the incident ray.

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