In Young's double-slit experiment, the intensity of light at a point on the screen where the path difference is `lambda` is `I, lambda` being the wavelength of light used. The intensity at a point where the path difference is `lambda//4` will be

Options

A. (a) `K`
B. (b) `K/4`
C. (c) `K/4`
D. (d) zero

Answer

C

Explanation

Phase difference, `phi=((2pi)/(lambda))` path difference
For path difference `lambda`, `phi=((2pi)/(lambda))lambda=2pi`
For path difference `lambda/4`, `phi=((2pi)/(lambda))lambda/4=pi/2`
Intensity on screen `I=4I_0cos^2(phi/4)`
For `phi=2pi, K=4I_0cos^2((2pi)/(4))impliesK=4I_0`
`I=Kco s^2 phi/2=Kcos2[(2pi)/(lambda)xxlambda/4xx1/2]`
`impliesKco s^2pi/4=K/2`
So intensity at given point where path difference is
`lambda/4` will be `K/2`.