NCERT Questions for Class 12 Physics Chapter 10 – Wave Optics

Important Questions with Solutions of Class 12 Physics Chapter 10 – Wave Optics

Short Answer Questions

1) The angular width of a fringe that forms on a distant screen in a double-slit experiment with light of wavelength 600 nm is 0.1^∘. How much distance separates the two slits?

Ans – Given that,

λ = 6000 nm = 600 × 10⁻⁹ m

The angular width of a fringe is correlated with slit spacing (d) as follows:

2) What is the Brewster angle for the transition from air to glass? We know refractive index of glass is 1.5.

Ans – Given that,

Refractive index μ = 1.5

Brewster angle = θ

Brewster angle and refractive index can be corelated as,

tan θ = μ

⇒ θ = tan^-1 1.5

⇒ θ = 56.31^∘

3) Distinguish between interference and diffraction.

Ans –

Interference	Diffraction
Interference arises from the superposition of light emitted by two coherent sources.	Diffraction results from the superposition of waves originating from several segments of the same wave front.
All dazzling fringes possess uniform intensity.	The intensity of bright fringes diminishes as the distance from the central light fringe increases.

4) The Doppler formula for frequency shift in sound waves varies significantly between the two scenarios presented, and this discrepancy warrants explanation. Do you anticipate the formulas to be entirely equivalent for the two scenarios involving light propagation in a medium?

a) Stationary source; moving observer, and

b) Source in motion; observer stationary. The exact Doppler equations for light waves in a vacuum are same in both scenarios.

Ans – Sound waves can only propagate through a medium. The two scenarios are not scientifically equivalent, as the velocity of an observer in relation to a medium differs between them. The Doppler formulae for the two specified scenarios cannot be identical.

In the case of light waves, sound cannot propagate in a vacuum. However, in a vacuum, both of these scenarios appear to be equivalent, as the speed of light remains unaffected by the observer’s motion or the velocity of the source. When light traverses a media, the aforementioned two examples are not equivalent, as the velocity of light is contingent upon the wavelength of the medium.

3 Mark Questions

1) Two coherent sources with an intensity ratio of 81:1 generate interference fringes. Determine the ratio of the intensities of peaks and minima in the interference pattern.

Ans – Given,

We know Intensity ∝ (Amplitude)²

2) What is the polarisation of light? Which types of waves exhibit the characteristic of polarisation? Identify two techniques for generating plane polarised light.

Ans – The phenomenon whereby the vibrations of a light vector are confined to a certain direction inside a plane perpendicular to the direction of light propagation is termed the polarisation of light. Transverse waves exhibit the characteristic of polarisation. There are two methods for generating plane polarised light:

• Polarisation by Reflection
• Scattering-induced polarisation.

3) An experiment using Young’s double slits repeats the slits at 0.24 mm. The screen is positioned 1.2 meters from the slits. The fringe width measures 0.03 cm. Determine the wavelength of the light utilised in the experiment.

Ans – Given,

β = 0.03 cm = 3 × 10⁻⁴ m

D = 1.2 m

d = 0.24 mm = 2.4 × 10⁻⁴ m

5 Marks Questions

1) Describe how corpuscular theory predicts that light will travel faster in a medium – such as water – than it will in a vacuum. Does the experimental measurement of the speed of light in water validate the prediction? If not, what other light image is compatible with the experiment?

Ans – No, wave theory is what it is.

According to Newton’s corpuscular theory of light, light particles encounter forces of attraction normal to the surface when they collide with the interface of two substances, one of which is denser (water) and the other is rarer (air). As a result, the component along the surface of velocity stays constant while the normal component grows.

Thus, we can write the following:

C sin i = v sin r … (1)

Where,
i = Angle of incidence
r = Angle of reflection
c = Velocity of light in air
v = Velocity of light in water

The relationship for the relative refractive index of water in relation to air is expressed as:

Equation (1) would therefore simplify to

However, μ is greater than 1. It can be inferred from equation (2) that v > c. This prediction is not feasible as it contradicts the experimental results of c > v. Consequently, the wave model of light aligns with the experimental findings.

2) A Young’s double-slit experiment is used to obtain interference fringes using a laser beam with two wavelengths, 650 nm and 520 nm.

(i) Determine the distance of the third bright fringe on the screen from the central maximum for a wavelength of 650 nm.

Ans – For n = 3 and λ = 650 nm

(ii) What is the minimum distance from the central maximum at which the bright fringes for both wavelengths coincide? λ₁ = 650 nm

Ans – Let the nth bright fringe corresponding to wavelength λ₂ overlap with the (n − 1)th bright fringe corresponding to wavelength λ₁ on the screen. The conditions for bright fringes can be equated as follows:

Therefore, the minimum distance from the centre maximum can be determined using the following relation:

3) (i) In a single slit diffraction experiment, the slit width is increased to double its initial dimension. What impact does this have on the core diffraction band’s size and intensity?

Ans – In a single slit diffraction experiment, if the slit width is doubled, the size of the centre diffraction band will decrease to half, while the strength of the central diffraction band will increase by a factor of four.

(ii) How does the interference pattern in a double-slit experiment relate to the diffraction from each slit?

Ans – The interference pattern in a double-slit experiment may be influenced by diffraction from each slit. The pattern would come from the interference of the diffracted waves from each slit.

(iii) When a small circular obstruction is positioned in the trajectory of light from a remote source, a luminous point is observed at the centre of the obstruction’s shadow. Elucidate the rationale.

Ans – When a small circular obstruction is positioned in the trajectory of light from a distant source, a luminous spot may be observed at the centre of the shadow cast by the obstruction. This occurs due to the diffraction of light waves from the edge of the circular obstacle, resulting in constructive interference at the middle of the shadow. This constructive interference would yield a luminous point.

(iv) Two students are divided by a 7 m partition wall in a room that is 10 m in height. Despite the ability of both light and sound waves to diffract around obstructions, how is it that the students cannot visually perceive one another while they can communicate audibly without difficulty?

Ans – The deflection of waves by barriers at a significant angle can occur when the obstacle’s size is comparable to the wavelength of the waves. The wavelength of the light waves is quite diminutive relative to the size of the obstruction. Consequently, the diffraction angle would be minimal. Consequently, the students would be unable to observe one another. The wall’s dimensions are comparable to the wavelength of the sound waves. Consequently, the waves would bend at a significant angle. Consequently, the students can hear one another.

(v) Ray optics is predicated on the premise that light propagates in a linear trajectory. Diffraction phenomena, discovered when light traverses narrow apertures or circumvents diminutive barriers, refute this premise. Nevertheless, the ray optics assumption is frequently employed to comprehend the positioning and other characteristics of images in optical devices. What is the rationale?

Ans – The rationale is that in conventional optical instruments, the aperture size is significantly bigger than the wavelength of the utilised light.