In an interference experiment the ratio of amplitudes of coherent waves is a _ 1 a _ 2 = (1)/(3) . The ratio of maximum and minimum intensities of fringes will be:

Correct answer: A. 4 I _ I _ = ( a _ 1 + a _ 2 a _ 1 - a _ 2 ) ^ 2 = ( a _ 1 / a _ 2 + 1 a _ 1 / a _ 2 - 1 ) ^ 2 = (((1)/(3) + 1)/((1)/(3) - 1)) ^ 2 I _ : I _ = 4: 1

JEE Main 20198 AprilMorning Shift

Exam: JEE Main
Year: 2019
Date: 8 April
Shift: Morning Shift
Subject: Physics
Chapter: Wave Optics

JEE Main 2019 · 8 April · Morning Shift · Physics · Wave Optics

In an interference experiment the ratio of amplitudes of coherent...

In an interference experiment the ratio of amplitudes of coherent waves is

\dfrac{{a}_{1}}{{a}_{2}} = \dfrac{1}{3}

. The ratio of maximum and minimum intensities of fringes will be:

Options

A $4$
B $9$
C $2$
D $18$

Correct answer

A. $4$

Step-by-step solution

\dfrac{{I}_\max}{{I}_\min} = {\left(\dfrac{{a}_{1} + {a}_{2}}{{a}_{1} - {a}_{2}}\right)}

^{2} = {\left(\dfrac{{a}_{1} / {a}_{2} + 1}{{a}_{1} / {a}_{2} - 1}\right)}^{2} = {\left(\dfrac{\dfrac{1}{3} + 1}{\dfrac{1}{3} - 1}\right)}^{2}

<br/>

{I}_\max: {I}_\min = 4: 1

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