In a Young's double slit experiment the intensity at a point where tha path difference is `(lamda)/(6)` (`lamda` being the wavelength of light used) is I. If `I_0` denotes the maximum intensity, `(I)/(I_0)` is equal to

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)`