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Q.

The angular deviation of 5th  order dark fringe is 12∘ in a single slit experiment. If the width of the slit is 9μm then the wavelength of the incident light is

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a

5892Ao

b

3768Ao

c

6022Ao

d

4862Ao

answer is D.

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Detailed Solution

5th order n=5

θ=120

We know that,  dsinθ=nλ  ,sinθ≈θ

⇒dθ=nλ

Given: θ=12×π180radius

 

λ=dθn=9×10−6×12×π180×5=0.3768×10−6

=3768A0

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