In Young's double slit experiment intensity of light emerging from each slit is I0. Wavelength of light used is λ, separation between the slits is d and separation between the slits and screen is D. P is a point on the screen where the resultant intensity is 3I0. Then the minimum path difference between rays emerging from the slits at point P is
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a
λ4
b
λ2
c
λ3
d
λ6
answer is D.
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Detailed Solution
At any point on the screen whose position is defined by the angle θ is I=Imcos2ϕ2=4I0cos2ϕ2At point P, 3I0=4I0cos2ϕ2⇒cosϕ2=32⇒ϕ=60o∴ ϕ=2πλ×path difference=π3⇒dsinθ=λ6