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Question

In a single slit diffraction experiment, light of wavelength 500 nm passes through a slit of width 2 μm. If the intensity of the central maximum is 40 Wm2, then the intensity of a point on a screen at an angular position of θ=30 is close to

A
0
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B
20 Wm2
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C
1 Wm2
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D
2 Wm2
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Solution

The correct option is A 0
Given:
λ=500 nm
a=2 μm

The intensity at a point on the screen is given by

I=I0(sinββ)2

Where,
I0Intensity at the central maximum

β=πasinθλ

Here, θ=30

β=π×2×106×(12)500×109

β=2π

I=I0(sin2π2π)2

I=0

Hence, option (A) is correct.
Alternate Method:

The condition for minimum in diffraction pattern is given by,

asinθ=nλ

2×106×(12)=n×500×109

n=2

Second minimum will be formed at the given angular position, hence, the intensity will be zero.

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