In the Young's double slit experiment, the intensity of light at a point on the screen where the path difference is `lambda` is K. (K being the wavelength of light used). The intensity at 2 a point where the path difference is `(lambda)/(4)` will be.....

Options

A. K
B. `(K)/(4)`
C. `(K)/(2)`
D. zero

Answer

C

Explanation

In `I=I_(0) cos^(2)[(K(r_(1)-r_(2)))/(2)],r_(1)-r_(2)=1 rArr I=K`
`:. K=I_(0) cos^(2)[(2pi)/(lambda)xx(lambda)/(2)]`
`:. K=I_(0)cos^(2) pi`
`:. K=I_(0)=....(I) " " [ :. Cos^(2)pi=(-1)^(2)=1]`
Now again `I.=I_(0)cos^(2)[(K(r_(1)-r_(2)))/(2)],r_(1)-r_(2)=(lambda)/(4)`
and `K=(2pi)/(lambda)`
`:.I.= I_(2)cos^(2)[(2pi)/(lambda)xx(lambda)/(8)]`
`=I_(0)cos^(2)(pi)/(4)`
`=(I_(0))/(2) [ :."cos"^(2)(pi)/(4)=((1)/(sqrt(2)))^(2)=(1)/(2)]`
`:. I=(K)/(2) [ :. I_(0)=` Kfrom equation (1)]