SATHEE: wave-optics Question 14

Wave Optics

wave-optics Question 14

Question: Q. 2. Find the intensity at a point on a screen in Young’s double slit experiment where the interfering waves have a path difference of (i) $λ / 6$ , and (ii) $λ / 2$ .

A [Foreign 2017]

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Solution:

Ans. Phase difference $= \frac{2 π}{λ} \times$ Path difference

Path difference $= \frac{λ}{6} \Rightarrow$ Phase difference $= \frac{π}{3} 1 / 2$

Path difference $\frac{λ}{2} \Rightarrow$ Phase difference $= π 1 / 2$

$I = 4 I_{0} \cos^{2} (\frac{ϕ}{2})$

(i)

$\begin{matrix} (ii) & I_{1} = 4 I_{0} \times \frac{3}{4} = 3 I_{0} \end{matrix}$

Detailed Answer :

Phase difference between interfering waves

$ϕ = \frac{2 π}{λ} \times path difference$

(i) Path difference $= \frac{λ}{6}$

$∴$ Phase difference $= \frac{2 π}{λ} \times \frac{λ}{6} = \frac{π}{3}$

$\begin{aligned} I & = 4 I_{0} \cos^{2} (\frac{ϕ}{2}) & = 4 I_{0} \cos^{2} \frac{π}{6} & = 4 I_{0} \times \frac{3}{4} = 3 I_{0} \end{aligned}$

(ii) Path difference $= \frac{λ}{2}$

$∴$ Phase difference, $ϕ = \frac{2 π}{λ} \times \frac{λ}{2} = π$

$\begin{aligned} I & = 4 I_{0} \cos^{2} \frac{ϕ}{2} & = 4 I_{0} \cos^{2} \frac{π}{2} = 0 \end{aligned}$

Wave Optics

wave-optics Question 14

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Wave Optics

wave-optics Question 14

Detailed Answer :

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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