NEET | 2016Wave OpticsEasy

Question

The intensity at the maximum in a Young’s double slit experiment is I0. Distance between two slits is d = 5λ, where λ is the wavelength of light used in the experiment. What will be the intensity in front of one of theslits on the screen placed at a distance D = 10 d?

Options

A.

I0

B.

I04

C.

34I0

D.

I02

Solution

In YDSE Imax = I0
Path difference at a point in front of one of shifts is
∆x=dyD=dd2D=d22D
∆x=d22(10 d)=d20=5λ20=λ4
Path difference is
ϕ=2πλ=(∆x)=2πλλ4
ϕ=π2
So intensity at that pt is
I = Imax cos2 θ2
I = I0 cos2 π4=I02

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