In young's double slit experiment, intensity of central maximum is Im. If wavelength of light used is 6000Ao, find the path difference between two waves emerging from the slits at a point on the screen where the intensity of light is Im/4.
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a
1000 Ao
b
4000 Ao
c
2000 Ao
d
3000 Ao
answer is C.
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Detailed Solution
I=Imcos2ϕ2⇒Im4=Imcos2ϕ2⇒cosϕ2=12⇒ϕ2=π3⇒ϕ=2π3Now 2πλ×path difference=ϕ⇒2π6000×path difference=2π3 ⇒path difference=2000 Ao