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Question

The intensity at the maximum in Young's double slit experiment is Io. The distance between two slits is d = 5 λ, where λ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance D = 10d?

A
Io
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B
Io4
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C
34Io
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D
Io2
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Solution

The correct option is D Io2
In the given, S1 & S2 are the two different slits.
Given,
d=5λ
Here, Imax=I0
Path difference=dY0D=d×d210λ=d20=λ4
[d=5λ]
For path difference λ4,Phase difference is d=2πλ×λ4=π2=90
We know, I=I0cos2902
I=I0(12)2
I=I02

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