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Question

In young double slit interference experiment using two coherent waves of different amplitudes, the intensity ratio between bright and dark fringes is 3. Then the value of the ratio of the amplitude of the waves that arrive there is:

A
[3+131]
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B
[313+1]
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C
3:1
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D
1:3
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Solution

The correct options are
A [313+1]
C [3+131]
Given : ImaxImin=3
Let the amplitude of the waves be A1 and A2 respectively.
Maximum intensity Imax=(I1+I2)2 where IiA2i
Imax=(A1+A2)2 ............(1)

Similarly, minimum intensity Imin=(A1A2)2
(A1+A2)2(A1A2)2=ImaxImin=3

OR A1+A2A1A2=3 A1=[3+131]A2

A1A2=[3+131] or, A2A1=[313+1]

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