Correct Answer - a
Consider light to be incident at near normal incidence. We wish to cause destructive interference between rays `r_(1)` and `r_(2)` so that maximum energy passes into the glass. A phase change ot `lambda // 2` occurs in each ray because at both the upper and lower surfaces of the `MgF_(2)` film the light is reflected by a medium of greater index of refraction, the light is reflected with no phase change. Since in this problem both rays 1 and 2 experience the same phase shift, no net change of phase is introduced by these two reflections. Hence, the only way a phase change can occur is if the two rays travel through different optical path lengths. The optical path length is the product of the geomaetric path difference a ray travels through different media and the refractive index of the medium in which it is travelling. For an odd number of half wavelength. Hence, the optical path difference needed for destructive interference is
`2 mu d = (2 n + 1) (lambda)/(2) , n = 0, 1, 2,...`
Now that `2 mu d` is the total optical path length that the rays traverse when `n = 0`. ltbRrgt `:. d = (lambda //2)/(2 mu) = (lambda)/(4 mu) = (350 xx 10^(-9))/(4 xx 1.38) = 100 nm = 1 xx 10^(-7) m`
