A right angled prism is to be made by selecting a proper material and the angles A and B (B

Q: Why does the incident ray enter undeviated?

The diagram indicates that the incident ray is perpendicular (normal) to the face $AB$. According to Snell's Law, a ray with an angle of incidence of $0^\circ$ will have an angle of refraction of $0^\circ$, meaning it passes straight through without bending.

Q: How do we know the angles of incidence are exactly $A$ and $B$?

Because the ray enters normally to $AB$, it is perpendicular to $AB$. The normal to face $AC$ is perpendicular to $AC$. The angle between the ray and the normal is the same as the angle between $AB$ and $AC$, which is $A$. A similar geometric argument applies at face $BC$.

Q: Why is face $BC$ the bottleneck for TIR?

For TIR to occur, the angle of incidence must be greater than the critical angle $\theta_c$. Since we are given $B \le A$, the angle of incidence at $BC$ (which is $B$) is smaller than at $AC$. If TIR happens at $BC$, it will definitely happen at $AC$.

Q: Why do we set $B = 45^\circ$ to find the minimum $\mu$?

A smaller refractive index $\mu$ results in a larger critical angle $\theta_c$. To satisfy $B \ge \theta_c$ with the largest possible $\theta_c$, we need the largest possible value for $B$. Since $A + B = 90^\circ$ and $B \le A$, the maximum value $B$ can take is $45^\circ$.

Q: In part (b), why can't we just change the angle of the incident ray?

The problem implicitly assumes the ray is incident normally, as shown by the perpendicular symbol in the original diagram. If the ray enters at a different angle, the symmetry is broken, and the specific conditions requested by the problem setup would change.

Question

A right angled prism is to be made by selecting a proper material and the angles $A$ and $B$ ($B \le A$), as shown in figure. It is desired that a ray of light incident on the face $AB$ emerges parallel to the incident direction after two internal reflections.
(a) What should be the minimum refractive index $\mu$ for this to be possible?
(b) For $\mu = \frac{5}{3}$ is it possible to achieve this with the angle $B$ equal to $30^\circ$ ?

Sigma Percentile · Accepted Answer

### The Bouncing Ray: Mastering Total Internal Reflection in a Prism

Imagine a ray of light entering a glass prism. If the conditions are just right, the light gets trapped inside, bouncing off the internal walls like a pinball before finally escaping. This phenomenon, known as Total Internal Reflection (TIR), is the secret behind optical fibers and brilliant diamond sparkles. Let's break down the geometry and physics of this fascinating problem.

### Analyzing the Setup

Look closely at the provided diagram. The incident ray strikes the hypotenuse $AB$ normally (perpendicularly). According to Snell's Law, a ray with an angle of incidence of $0^\circ$ will have an angle of refraction of $0^\circ$. This means it enters the prism without any deviation.

Once inside, it travels straight until it hits face $AC$ at point $P$. It reflects off $AC$, travels to face $BC$, hits it at point $Q$, and reflects again. Finally, it strikes face $AB$ from the inside and emerges parallel to its original direction.

### The Master Equation: Angles of Incidence

To understand if the ray will reflect or escape at faces $AC$ and $BC$, we need to find the angles of incidence at these points.

If we draw a normal at $P$ (which is perpendicular to $AC$), simple geometry reveals that the angle the ray makes with this normal is exactly equal to the prism angle $A$. Therefore, the angle of incidence at $P$ is:

$$i_A = A$$

Similarly, when the ray hits face $BC$ at $Q$, its angle of incidence is exactly equal to the prism angle $B$:

$$i_B = B$$

### The Bottleneck for TIR

For the ray to stay inside and eventually emerge back from face $AB$, it must undergo Total Internal Reflection at **both** faces $AC$ and $BC$. This means both angles of incidence must be greater than or equal to the critical angle $	heta_c$ of the prism material.

$$i_A \ge 	heta_c \quad 	ext{and} \quad i_B \ge 	heta_c$$

We are given that angle $B \le A$. Consequently, $i_B \le i_A$. The smaller angle of incidence is at $Q$. So, if TIR happens at $Q$, it will definitely happen at $P$. Our bottleneck condition is simply:

$$B \ge 	heta_c$$

### Final Calculation: Minimum Refractive Index

We want to find the minimum possible refractive index $\mu$. A smaller $\mu$ means a larger critical angle $	heta_c$. To accommodate the largest possible critical angle, we need the largest possible value for angle $B$.

In a right-angled triangle, $A + B = 90^\circ$. Since $B$ cannot exceed $A$, the maximum value $B$ can take is exactly $45^\circ$ (when $A = B = 45^\circ$).

Let's substitute this maximum value into our bottleneck condition:

$$45^\circ \ge 	heta_c$$

Taking the sine on both sides:

$$\sin 45^\circ \ge \sin 	heta_c$$

We know that $\sin 	heta_c = \frac{1}{\mu}$. Substituting this in:

$$\frac{1}{\sqrt{2}} \ge \frac{1}{\mu}$$

Rearranging this inequality, we find:

**$$\mu \ge \sqrt{2}$$**

So, the minimum refractive index required is $\sqrt{2}$.

### Part (b): Testing Specific Values

Now let's tackle part (b). We are given a specific refractive index, $\mu = \frac{5}{3}$, and a specific angle $B = 30^\circ$.

First, let's calculate the critical angle for this material:

$$\sin 	heta_c = \frac{1}{\mu} = \frac{3}{5}$$

This corresponds to a critical angle of approximately $37^\circ$.

Our angle of incidence at face $BC$ is equal to angle $B$, which is $30^\circ$. But wait, $30^\circ$ is strictly less than our critical angle of $37^\circ$!

Because the angle of incidence is too small ($i_B < 	heta_c$), Total Internal Reflection will fail at face $BC$, and the light will escape the prism prematurely. Therefore, it is **not possible** to achieve the desired ray path with these values.

Visualized Solution

$Analyzing the Ray Path$

$Angles of Incidence$

$Condition for TIR$

$Maximizing B$

$Minimum Refractive Index$

$Checking for μ = \frac{5}{3} and B = 3 0^{\circ}$

$Conclusion for Part (b)$

The Sigma Insight: Refraction and Total Internal Reflection

The Bouncing Ray

Analyzing the Setup

The Master Equation

The Bottleneck for TIR

Final Calculation

Part (b)

Similar Questions

A parallel beam of light is incident from air at an angle $α$ on the side $P Q$ of a right angled triangular prism of refractive index $n = 2$ . Light undergoes total internal reflection in the prism at the face $P R$ when $α$ has a minimum value of $4 5^{\circ}$ . The angle $θ$ of the prism is

A light ray is incident perpendicular to one face of a $9 0^{\circ}$ prism and is totally internally reflected at the glass-air interface. If the angle of reflection is $4 5^{\circ}$ , we conclude that for the refractive index $n$ as

A ray $O P$ of monochromatic light is incident on the face $A B$ of prism $A B C D$ near vertex $B$ at an incident angle of $6 0^{\circ}$ (see figure). If the refractive index of the material of the prism is $3$ , which of the following is (are) correct?

A glass prism of refractive index 1.5 is immersed in water (refractive index 4/3). A light beam incident normally on the face $A B$ is totally reflected to reach the face $B C$ if

A ray of light passes through four transparent media with refractive indices $μ_{1}$ , $μ_{2}$ , $μ_{3}$ and $μ_{4}$ as shown in the figure. The surfaces of all media are parallel. If the emergent ray $C D$ is parallel to the incident ray $A B$ , we must have

Light is incident at an angle $α$ on one planar end of a transparent cylindrical rod of refractive index $n$ . Determine the least value of $n$ so that the light entering the rod does not emerge from the curved surface of the rod irrespective of the value of $α$ .

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing the Ray Path

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Angles of Incidence

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for TIR

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Maximizing B

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Minimum Refractive Index

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Checking for μ=35​ and B=30∘

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion for Part (b)