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

Options

A. `(1)/(sqrt(2))`
B. `(sqrt(3))/(2)`
C. `(1)/(2)`
D. `(3)/(4)`

Answer

D

Explanation

The intensity at a general point with respect to maximum intensity is `I=I_(0)cos^(2)((phi)/(2))`
Phase difference `phi=((2pi)/(lamda))` (Path difference)
`impliesphi=(2pi)/(lamda)(lamda)/(6)=(pi)/(3)`
Hence, `(I)/(I_(0))=[cos((60)/(2))]^(2)=((sqrt(3))/(2))^(2)=(3)/(4)`