In a double – slit experiment, at a certain point on the screen the path difference between the two interfering waves is (1)/(8)th of a wavelength. The ratio of the intensity of light at that point to that at the center

Correct answer: A. 0.853 I = I _ 0 ^ 2 ((∆ ϕ)/(2)) I I _ 0 = ^ 2 [((2π)/(λ) × ∆ X)/(2)] = ^ 2 ((π)/(8)); I I _ 0 = 0.853

JEE Main 20208 JanuaryEvening Shift

Exam: JEE Main
Year: 2020
Date: 8 January
Shift: Evening Shift
Subject: Physics
Chapter: Wave Optics

JEE Main 2020 · 8 January · Evening Shift · Physics · Wave Optics

In a double – slit experiment, at a certain point on the screen the...

In a double – slit experiment, at a certain point on the screen the path difference between the two interfering waves is

\dfrac{1}{8}th

of a wavelength. The ratio of the intensity of light at that point to that at the center of a bright fringe is:

Options

A $0.853$
B $0.672$
C $0.568$
D $0.760$

Correct answer

A. $0.853$

Step-by-step solution

$I = {I}_{0}\cos^{2}\left(\dfrac{∆ ϕ}{2}\right)$
$\dfrac{I}{{I}_{0}} = \cos^{2}\left[\dfrac{\dfrac{2\pi}{\lambda} \times ∆ X}{2}\right] = \cos^{2}\left(\dfrac{\pi}{8}\right); \dfrac{I}{{I}_{0}} = 0.853$

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