Question

When reaching a boundary between two media (1 and 2), an incident ray is partially reflected and partially refracted. The ray is travelling from media 1 to media 2.At what angle of incidence is the reflected ray perpendicular to the incident ray? The indexes of refraction for the two media are n1 and n2, respectively.

Answer 1

Answer:

The angle of incidence when the reflected ray is perpendicular to the incident ray = 45°

Explanation:

According to Snell's Law,

n₁ sin θ₁ = n₂ sin θ₂

When the angle between the incident ray and reflected ray is 90°, the angle of incidence is θ₁ and the angle of reflection, θ₂ = 90° - θ₁ and the index of refraction in the Snell's Law for both media would be the same, n₁ = n₂ = n

n sin θ₁ = n sin (90° - θ₁)

Note that from trigonometric relations,

Sin (90° - θ₁) = cos θ₁

n sin θ₁ = n cos θ₁

(sin θ₁)/(cos θ₁) = 1

tan θ₁ = 1

θ₁ = arctan 1 = 45°

Hope this Helps!!!

Answer 2

Answer:

The angle of incidence is $\theta _{B}=tan^{-1} \frac{n_{2} }{n_{1} }$

Explanation:

For a p-polarized light:

$r_{12p} =\frac{tan(\theta _{1} -\theta _{2} )}{tan(\theta _{1} +\theta _{2} ) }$

Where

r₁₂p = Fresnel reflection coefficient for p-polarized

The same way, for a s-polarized light:

$r_{12s} =\frac{sin(\theta _{1} -\theta _{2} )}{sin(\theta _{1} +\theta _{2} ) }$

Where

r₁₂s = Fresnel reflection coefficient for s-polarized

If the light is reflected, then there will have a s-polarization. The incident angle (Brewster angle) is equal to:

$n_{1} sin\theta _{B} =n_{2} sin((n/2)-\theta _{B})=n_{2} cos\theta _{B}\\tan\theta _{B}=\frac{n_{2} }{n_{1} } \\\theta _{B}=tan^{-1} \frac{n_{2} }{n_{1} }$