[ف.نووي]

A light ray is partially reflected and partially refracted at a boundary between two media, the upper one having index of refraction n, the lower one having index of refraction n, as shown in the figure. The reflected ray is perpendicular to the refracted ray when

a. n' = n tanθ incident.
b. n' = n cot θincident.
c. n' = n .
d. n' = n .
e. n' = n sec θincident.
Solution:

If .. θ_r+ θ_i⁡= 90 >>>>> 1

Applying the law of refraction in this case::
n sin( θ_i) = n' sin (θ_r) >>>>> 2

Compensation of 1 in 2, we find that::

n sin( θ_i) = n' sin (90 - θ_i) >>>>> 3

It is well known that:

(sin (90 - θ_i) = cos (θ_i))

If you become a 3 on the following image::

n sin( θ_i) = n' cos (θ_i)

n'/n=(sinθ_i)/(cosθ_i )=tan⁡(θ_i
n'=n tan⁡(θ_i