The intensity at the maximum in a Young's double slit experiment is `I_0`. Distance between two slits is `d=5lambda`, where `lambda` is the wavelength of light used in the experiement. What will be the intensity in front of one of the slits on the screen placed at a distance `D=10d`?

Options

A. (a) `I_0/2`
B. (b) `3/4I_0`
C. (c) `I_0`
D. (d) `I_0/4`

Answer

A

Explanation

Suppose P is a point in front of one slit at which `x=d/2`. Path difference between the waves reaching at P
`d=(xd)/(D)=((d/2)d)/(10d)=(d)/(20)=(5lambda)/(20)=lambda/4`

Hence corresponding phase difference
`phi=(2pi)/(lambda)xx(lambda)/(4)=lambda/2`
Resultant intensity at P
`I=I_(max)cos^2phi/2=Icos^2(pi/4)=I_0/2`