SATHEE: wave-optics Question 1

wave-optics Question 1

Question: Q. 1. Consider a ray of light incident from air onto a slab of glass (refractive index $n$ ) of width $d$ , at an angle $θ$ . The phase difference between the ray reflected by the top surface of the glass and the bottom surface is

(a) $\frac{4 π d}{λ} {(1 - \frac{1}{n^{2}} \sin^{2} θ)}^{\frac{1}{2}} + π$ .

(b) $\frac{4 π d}{λ} {(1 - \frac{1}{n^{2}} \sin^{2} θ)}^{\frac{1}{2}}$ .

(d) $\frac{4 π d}{λ} {(1 - \frac{1}{n^{2}} \sin^{2} θ)}^{\frac{1}{2}} + 2 π$ .

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Solution:

Ans. Correct option : (a)

Explanation: If slab of a glass is placed in air, the wave reflected from the upper surface (from a denser medium) suffers a sudden phase change of $π$ , while the wave reflected from the lower surface (from a rarer medium) suffers no such phase change.

Now consider the diagram, the ray $(P O)$ is incident at an angle $θ$ and gets reflected in the direction $P^{'}$ and refracted in the direction $P^{''}$ . Due to reflection from the glass medium, there is a phase change of $π$ .

Time taken to travel along $O P^{''}$ ,

$Δ t = \frac{O P^{''}}{v} = \frac{\frac{d}{\cos r}}{\frac{c}{n}} = \frac{n d}{c \times \cos r}$

From Snell’s law, $n = \frac{\sin θ}{\sin r}$

$\begin{aligned} \sin r & = \frac{\sin θ}{n} \cos r & = \sqrt{1 - \sin^{2} r} & = \sqrt{1 - \frac{\sin^{2} θ}{n^{2}}} \end{aligned}$

Phase difference,

$\begin{aligned} Δ ϕ & = \frac{2 π}{T} \times Δ t \Rightarrow Δ ϕ & = \frac{2 π n d}{λ} {(1 - \frac{\sin^{2} θ}{n^{2}})}^{\frac{- 1}{2}} \end{aligned}$

So, net phase difference $= Δ f + π$

$\Rightarrow Δ ϕ_{net} = \frac{4 π d}{λ} {(1 - \frac{1}{n^{2}} \sin^{2} θ)}^{\frac{1}{2}} + π$

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Wave Optics

wave-optics Question 1

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Wave Optics

wave-optics Question 1

Very Short Answer Type Questions

Table of Contents

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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