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:
The angle of incidence is [tex]\theta _{B}=tan^{-1} \frac{n_{2} }{n_{1} }[/tex]
Explanation:
For a p-polarized light:
[tex]r_{12p} =\frac{tan(\theta _{1} -\theta _{2} )}{tan(\theta _{1} +\theta _{2} ) }[/tex]
Where
r₁₂p = Fresnel reflection coefficient for p-polarized
The same way, for a s-polarized light:
[tex]r_{12s} =\frac{sin(\theta _{1} -\theta _{2} )}{sin(\theta _{1} +\theta _{2} ) }[/tex]
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:
[tex]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} }[/tex]
