Q.Show that for a material with refractive index $\mu \geq \sqrt{2}$, light incident at any angle shall be guided along a length perpendicular to the incident face.

Concept: Total Internal Reflection (TIR) — light is confined when the angle of incidence inside the medium exceeds the critical angle.

Reasoning:

1. Light enters the material from air ($\mu_{\text{air}} = 1$) at an angle of incidence $i$. By Snell’s law:

   $\sin i = \mu \sin r$, where $r$ is the angle of refraction inside the material.

2. The refracted ray strikes the perpendicular face (side wall) at an angle $\theta = 90^\circ - r$. For TIR at that wall, we need $\theta \geq \theta_c$, where $\sin \theta_c = 1/\mu$.

3. So the condition becomes:

   $\sin(90^\circ - r) = \cos r \geq \frac{1}{\mu}$.

   Using $\cos r = \sqrt{1 - \sin^2 r} = \sqrt{1 - \frac{\sin^2 i}{\mu^2}}$, the worst case is grazing incidence ($i \to 90^\circ$, $\sin i = 1$). …

The key insight here is Total Internal Reflection (TIR). When light travels from a denser medium (the material, refractive index $\mu$) to a rarer medium (air, refractive index $1$), it bends away from the normal. If the angle of incidence inside the material exceeds a certain critical angle $i_c$, the ray cannot escape — it reflects back entirely, like a mirror.

The problem asks: when does any ray entering one face get trapped and guided along a perpendicular direction? That means the ray must hit the adjacent face at an angle greater than $i_c$, no matter how it entered.

Let’s set up the geometry.

1. Define the path. Imagine a rectangular slab. Light enters through the left face (the incident face). Inside the material, it bends toward the normal (since $\mu > 1$). It then travels to the top face (perpendicular to the incident face). For the ray to be guided along the length, it must reflect off this top face via TIR — so its angle of incidence at the top face must be $\geq i_c$.

2. Relate the angles. Let the angle of incidence at the left face be $i$ (measured from the normal to that face). Inside the material, the ray makes an angle $r$ with the normal, given by Snell’s law:

Now, look at the top face. The normal to the top face is perpendicular to the normal of the left face. So the angle the ray makes with the top face’s normal is $90^\circ - r$. Call this angle $\theta$:

For TIR at the top face, we need $\theta \geq i_c$, where $i_c$ is the critical angle for the material-air interface:

3. Translate the condition. The condition $\theta \geq i_c$ becomes:

Taking sines (since all angles here are between $0^\circ$ and $90^\circ$, sine is increasing):

But $\cos i_c = \sqrt{1 - \sin^2 i_c} = \sqrt{1 - \frac{1}{\mu^2}}$.

4. Substitute for $\sin r$. From Snell’s law, $\sin r = \frac{\sin i}{\mu}$. So the TIR condition becomes:

Multiply through by $\mu$:

$\sin i \leq \mu \sqrt{1 - \frac{1}{\mu^2}} = \sqrt{\mu^2 - 1}.$ …

Method: Proving a Total-Internal-Reflection Condition Holds for "Any" Angle of Incidence

Use this method when a question asks you to show that TIR occurs at a second (perpendicular, or otherwise related) face regardless of the angle at which light first enters — i.e. to find the minimum refractive index that guarantees guiding for every possible incident ray.

Steps

Step 1: Relate the entry angle to the angle at the second face

Use Snell's law at the entry face to express the refraction angle $r$ (inside the medium) in terms of the incidence angle $i$: $\sin r = \dfrac{\sin i}{\mu}$. Then use the geometry of the shape (e.g. two faces at $90^\circ$ to each other) to express the angle $\theta$ at which the ray strikes the second face in terms of $r$ — typically $\theta = 90^\circ - r$ for perpendicular faces.

Step 2: Write down the TIR condition at the second face

For the ray to be totally internally reflected there (and thus guided, not escape), require $\theta \geq \theta_c$, the critical angle for that interface, where $\sin\theta_c = 1/\mu$.

Step 3: Combine the two relations into a single inequality in $i$ and $\mu$ …