Light of wavelength `lambda_(0)` in air enters a medium of refractive index n. If two points A and B in this medium lie along the path of this light at a distance x, then phase difference `phi_(0)` between these two point is

Options

A. `phi_(0) = (1)/(n) ((2 pi)/(lambda_(0))) x`
B. `phi_(0) = n ((2 pi)/(lambda_(0))) x`
C. `phi_(0) = (n - 1) ((2 pi)/(lambda_(0))) x`
D. `phi_(0) = (1)/((n - 1)) ((2 pi)/(lambda_(0))) x`

Answer

b

Explanation

`q phi_(0) = (2 pi)/(lambda_(0)) ("optical path")`
`phi_(0) = (2 pi)/(lambda_(0)) (nx)`