Umakant Kondapure, Collin Fernandes, Nipun Bhatia, Vikram Bathula and, Ketki Deshpande Solutions for Chapter: Wave Theory of Light, Exercise 2: Critical Thinking

Author:Umakant Kondapure, Collin Fernandes, Nipun Bhatia, Vikram Bathula & Ketki Deshpande

Umakant Kondapure Physics Solutions for Exercise - Umakant Kondapure, Collin Fernandes, Nipun Bhatia, Vikram Bathula and, Ketki Deshpande Solutions for Chapter: Wave Theory of Light, Exercise 2: Critical Thinking

Attempt the practice questions on Chapter 4: Wave Theory of Light, Exercise 2: Critical Thinking with hints and solutions to strengthen your understanding. MHT-CET TRIUMPH Physics Multiple Choice Questions Part - 2 Based on Std. XI & XII Syllabus of MHT-CET solutions are prepared by Experienced Embibe Experts.

Questions from Umakant Kondapure, Collin Fernandes, Nipun Bhatia, Vikram Bathula and, Ketki Deshpande Solutions for Chapter: Wave Theory of Light, Exercise 2: Critical Thinking with Hints & Solutions

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A plane glass slab is kept over various coloured letters. The letter which appears the least raised is,

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A ray of light is incident normally on a glass slab of refractive index $\mu$ of thickness $d$ which is at a distance $x$ from the glass. The ray of light takes same time to reach from source to slab and to pass through the slab. The thickness of the slab is,

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Light entering an air glass $(\mu=1.5)$ boundary is partly reflected and partly refracted. If the incident and reflected rays are at right angles to each other, the angle of refraction $r$ is given by,

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A plane mirror PQ is held normally to water surface of refractive index 1.33. A ray of light is incident at an angle of 50° with the mirror surface. After reflection, the ray is refracted into water. The angle of refraction r is, here sin50°=0.760.

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A ray of light strikes a glass plate at an angle of 60°. If the reflected and refracted rays are perpendicular to each other, the index of refraction of glass is,

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A light source approaches the observer with velocity $0.8 \mathrm{c}$. The Doppler shift for the light of wavelength 5500 Å is,

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Assertion: Contribution of the wavelets lying on back on a wavefront is zero.

Reason: The contribution of a wavelet in any direction making angle $\theta$ with the wavelet is proportional to $\frac{1}{2}(1+\cos \theta)$. In this case, θ=180°.

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Assertion: The colour of the light can be assessed from the wavelength of light waves.

Reason: Intensity $=(\text { Amplitude })^{2}$.