In Young's double slit experiment intensity at a point is `((1)/(4))` of the maximum intersity. Angular position of this point is

Options

A. `sin^(-1) (lambda // d)`
B. `sin^(-1) (lambda // 2d)`
C. `sin^(-1) (lambda // 3d)`
D. `sin^(-1) (lambda // 4d)`

Answer

c

Explanation

Let P be the point on the central maxmia whose intensity is one-fourth of the maximum intensity.
For interference we know that `I = I_(1) + I_(2) + 2 sqrt(I_(1) I_(2)) cos phi`, where I is the intensity at P, `I_(1), I_(2)` are the intensities of light origination from A and B, respectively, and `phi` is the phase diffenence at P. In YDSE, `I_(1) = I_(2) = I` and `I_(max) = 4 I`. We are concentrating at a point where the intensity is one-fourth of the maxima intensity.
`:. I = I + I + 2I cos phi`
`implies -(1)/(2) = cos phi implies phi = (2pi)/(3)`
Note that we the least value of the angle as the point is in central maxima.
For a phase difference of `2pi`, the path difference is `lambda`.
For a phase of `(2 pi)/(3)` , the
path difference is `(lambda)/(2 pi) xx (2 pi)/(3) = (lambda)/(3)`.
But the path difference (in terms of P and Q) is `theta` as shown in figure.
`:. d sin theta =(lambda)/(3)`
`implies sin theta = (lambda)/(3d) implies theta = sin^(-1) ((lambda)/(3d))`
Hence, (c) is the correct option.