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

Answer

A

Explanation

As `I_(R ) = I_(1) + I_(2) + 2sqrt(I_(1) I_(2)) cos phi`
`:. I_(0) = I_(max) = I + I + 2sqrt(I xx I) cos 0^(@) = 4 I`
For path differenc, `lambda//6`
phase difference, `theta = (360^(@))/(6) = 60^(@)`
`:. I' = I + I + 2sqrt(I I) cos 60^(@) = 3 I`
`(I')/(I_(0)) = (3 I)/(4 I) = (3)/(4)`