Question

Suppose you are using total internal reflection to make an efficient corner reflector. If there is air outside and the incident angle is 45.0º , what must be the minimum index of refraction of the material from which the reflector is made?

Short answer

As a result, the minimum refractive index of the material used to make the reflector is 1.414.

Step-by-step solution

Definition of angle of incident

At the point of incidence, a line establishes an angle with a line perpendicular to the surface.

Given data and formula used

• Critical angle, ${\theta }_{\circ }={45}^{°}$
• Refractive index of air, n₁=1
• The expression to find the critical angle is,

$\sin {\theta }_{\circ }=\frac{n_1}{n_2}........\left(1\right)$

Here,

• ${\theta }_{\circ }$is critical angle
• $n_1$is the refractive index of air.
• $n_2$is the refractive index of the material of the reflector.

Finding the Minimum index of refraction of the material

Let us find the minimum index of refraction of the material.

Rearranging the equation ( 1 ),

$n_2=\frac{n_1}{\sin {\theta }_{\circ }}..........\left(2\right)$

Substituting the value of n₁ by 1 and${\theta }_{\circ }={45}^{°}$ in equation ( 2 ),

$$\begin{gathered} n_2=\frac{1}{\sin {45}^{°}} \\ =\frac{1}{{\displaystyle \frac{1}{\sqrt{2}}}} \\ =\sqrt{2} \\ =1.414 \end{gathered}$$

Therefore, the material utilized to build the reflector has a minimum refractive index of 1.414.