For the given incident ray as shown in figure, the condition of total internal reflection of the ray will be satisfied if the refractive indesx of block will be

Options

A. `cos theta=3lambda//2d`
B. `cos theta=lambda//4d`
C. `sec theta-cos theta=lambda//d`
D. `sec theta-cos theta=4lambda//d`

Answer

B

Explanation

Path difference between the two rays is given by

`Delta=CO+PO`
`because PR=d`
So, `PO=dsec thetha`
and `CO=PO cos 2theta=dsec thetacos 2theta`
So, `Delta=(d sec theta+d sec theta cos 2 theta)`
Phase difference between two rays is `phi=pi` (as one ray is reflected one and another is direct)
Now, for constructive interference path difference should be even multiple of half wavelength
i.e., `Delta=lambda//2, 3lambda//2`
So, `d sec theta+d sec theta cos 2theta=lambda/2`
or `d sec theta(1+cos 2theta)=lambda/2`
or `d sec theta(2cos^(2)theta)=lambda/2`
`:. cos theta=lambda//4d`