The maximum intensity in Young's double slit experiment is I0. Distance between the slits is d=5λ, where λ is the wavelength of monochromatic light used in the experiment. What will be the intensity of light in front of one of the slits on a screen at a distance D = d ?
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a
I02
b
34I0
c
I0
d
I04
answer is A.
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Detailed Solution
Path difference Δx=ydD here y=ydD ( as d=5λ)and D=10d=50λ so Δx=5λ25λ50λ=λ4Corresponding phase difference will beϕ=2πλ(Δx)=2πλλ4=π2 or ϕ2=π4⇒I=I0cos2ϕ2=I0cos2π4=I02
The maximum intensity in Young's double slit experiment is I0. Distance between the slits is d=5λ, where λ is the wavelength of monochromatic light used in the experiment. What will be the intensity of light in front of one of the slits on a screen at a distance D = d ?