A right angled prism (45°-90°-45°) of refractive index n has a plane of refractive index n1 (n1

Q: Why does the ray graze the surface for the critical angle?

At the critical angle, the angle of refraction becomes $90^\circ$. This means the refracted ray travels exactly along the boundary between the two media.

Q: How do we know $r_1 + r_2 = A$?

In any prism, the sum of the internal angles of refraction at the two faces equals the apex angle of the prism. This can be proven using the quadrilateral formed by the two normals and the prism apex.

Q: Why is the normal at face AC drawn at $45^\circ$?

The prism is a right-angled isosceles prism with angles $45^\circ$-$90^\circ$-$45^\circ$. The face AB is vertical, and BC is horizontal. The face AC is the hypotenuse, inclined at $45^\circ$. Therefore, its normal is also inclined at $45^\circ$ to the horizontal.

Q: What happens if $n_1 > n$?

If the cemented layer has a higher refractive index than the prism, total internal reflection cannot occur at the AC interface. The ray would always refract into the cemented layer.

Q: Why does the ray pass undeviated in part (b)?

When a ray strikes a surface normally (perpendicularly), the angle of incidence is $0^\circ$. According to Snell's law, the angle of refraction will also be $0^\circ$, meaning the ray continues in the same straight line without bending.

Question

A right angled prism ($45^\circ$-$90^\circ$-$45^\circ$) of refractive index $n$ has a plane of refractive index $n_1$ ($n_1 < n$) cemented to its diagonal face. The assembly is in air. The ray is incident on $AB$.
(a) Calculate the angle of incidence at $AB$ for which the ray strikes the diagonal face at the critical angle.
(b) Assuming $n = 1.352$, calculate the angle of incidence at $AB$ for which the refracted ray passes through the diagonal face undeviated.

Sigma Percentile · Accepted Answer

This problem is a beautiful exploration of how light navigates through a prism, specifically focusing on the boundary conditions of total internal reflection and normal incidence. Let's break down the journey of the light ray step by step.

### Analyzing the Setup

We are given a right-angled isosceles prism with angles $45^\circ$-$90^\circ$-$45^\circ$. The refractive index of the prism is $n$. A plane layer of a different material with a lower refractive index $n_1$ (where $n_1 < n$) is cemented to the diagonal face $AC$. The entire assembly is placed in air.

When a light ray enters the prism through the vertical face $AB$, it refracts and bends towards the normal. Let the angle of incidence at $AB$ be $i_1$ and the angle of refraction be $r_1$. The ray then travels through the prism and strikes the diagonal face $AC$ at an angle of incidence $r_2$.

### Part (a): The Critical Angle Condition

For part (a), we are asked to find the angle of incidence $i_1$ such that the ray strikes the face $AC$ exactly at the critical angle. When a ray strikes an interface at the critical angle, it grazes along the boundary.

The critical angle $	heta_c$ at the interface between the prism (refractive index $n$) and the cemented layer (refractive index $n_1$) is given by:

$$ \sin 	heta_c = \frac{n_1}{n} $$

From the geometry of any prism, the sum of the internal angles of refraction equals the apex angle of the prism. Here, the apex angle $A$ is $45^\circ$. Therefore:

$$ r_1 + r_2 = A = 45^\circ $$

Since we want the ray to strike $AC$ at the critical angle, we set $r_2 = 	heta_c$. This allows us to express $r_1$ as:

$$ r_1 = 45^\circ - 	heta_c $$

### The Master Equation

Now, we apply Snell's law at the first interface $AB$. The ray is coming from air (refractive index $1$) into the prism:

$$ 1 \cdot \sin i_1 = n \sin r_1 $$

Substituting our expression for $r_1$:

$$ \sin i_1 = n \sin (45^\circ - 	heta_c) $$

To solve this, we expand the sine term using the trigonometric identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$:

$$ \sin i_1 = n (\sin 45^\circ \cos 	heta_c - \cos 45^\circ \sin 	heta_c) $$

Since $\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}$, we can factor it out:

$$ \sin i_1 = \frac{n}{\sqrt{2}} (\cos 	heta_c - \sin 	heta_c) $$

We already know $\sin 	heta_c = \frac{n_1}{n}$. Using the Pythagorean identity, we can find $\cos 	heta_c$:

$$ \cos 	heta_c = \sqrt{1 - \sin^2 	heta_c} = \sqrt{1 - \frac{n_1^2}{n^2}} = \frac{\sqrt{n^2 - n_1^2}}{n} $$

Substituting these into our expanded Snell's law equation:

$$ \sin i_1 = \frac{n}{\sqrt{2}} \left( \frac{\sqrt{n^2 - n_1^2}}{n} - \frac{n_1}{n} ight) $$

Notice how elegantly the $n$ in the numerator cancels with the $n$ in the denominators! We are left with a clean, final expression:

$$ \sin i_1 = \frac{1}{\sqrt{2}} (\sqrt{n^2 - n_1^2} - n_1) $$

Taking the inverse sine gives us the required angle of incidence for part (a):

$$ i_1 = \sin^{-1} \left\{ \frac{1}{\sqrt{2}} (\sqrt{n^2 - n_1^2} - n_1) ight\} $$

### Part (b): The Undeviated Ray

For part (b), the condition changes. We are told that the ray passes undeviated through the diagonal face $AC$. For a ray to pass through an interface without bending, it must strike the surface normally (perpendicularly). This means the angle of incidence at $AC$ is zero:

$$ r_2 = 0^\circ $$

Using our geometric relation $r_1 + r_2 = 45^\circ$ again, we find:

$$ r_1 = 45^\circ - 0^\circ = 45^\circ $$

Now, we simply apply Snell's law at face $AB$ one more time, using the given refractive index $n = 1.352$:

$$ \sin i_1 = n \sin r_1 $$
$$ \sin i_1 = 1.352 	imes \sin 45^\circ $$
$$ \sin i_1 = 1.352 	imes \frac{1}{\sqrt{2}} \approx 0.956 $$

Taking the inverse sine of $0.956$, we find the angle of incidence:

$$ i_1 = \sin^{-1}(0.956) \approx 73^\circ $$

And there we have it! By carefully tracking the geometry and applying Snell's law at each boundary, we've successfully navigated the light ray through the prism for both scenarios.

Visualized Solution

\text{Visualizing the Setup}

\text{Critical Angle at Face } AC

\text{Geometry of the Prism}

\text{Snell's Law at Face } AB

\text{Expanding the Sine Term}

\text{Substituting } \theta_c

\text{Final Answer for Part (a)}

\text{Condition for Undeviated Ray}

\text{Calculating } r_1 \text{ for Part (b)}

\text{Final Calculation for Part (b)}

The Sigma Insight: Refraction and Total Internal Reflection

Analyzing the Setup

Part (a)

The Master Equation

Part (b)

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