A transparent block A having refractive index μ = 1.25 is surrounded by another medium of refractive index μ = 1.0 as shown in figure. A light ray is incident on the flat face of the block with incident angle θ as shown in figure. What is the maximum value of θ for which light suffers total internal reflection at the top surface of the block?

 

❓ Question

A transparent block A having refractive index $\mu = 1.25$ is surrounded by another medium of refractive index $\mu = 1.0$ (air). A light ray is incident on the flat vertical face of the block with incident angle $\theta$ (in air) as shown in the figure. What is the maximum value of $\theta$ for which the light suffers total internal reflection at the top horizontal surface of the block?

(Assume the top surface is horizontal and the vertical face is flat — standard rectangular block geometry.)

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🖼️ Question Image

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✍️ Short Solution

Step 1 — Snell’s law at the vertical face

Air $(n_1=1.0)$ → block $(n_2=1.25)$.
At the vertical face:

$$n_1\sin \theta =n_2\sin r\Rightarrow \sin r=\frac{n_1}{n_2}\sin \theta =\frac{1}{1.25}\sin \theta =0.8\sin \theta \mathrm{.}$$

Here $r$ is the angle the refracted ray makes with the normal to the vertical face (i.e. with the horizontal direction).

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Step 2 — Angle of incidence at the top (horizontal) surface

The top surface is horizontal, so its normal is vertical. The angle of incidence $\phi$ (inside the block) on the top surface equals the angle between the ray and the vertical normal.

Since the ray makes angle $r$ with the horizontal, its angle with the vertical is

$$\varphi ={90}^{\circ }-r\mathrm{.}$$

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Step 3 — Condition for total internal reflection (TIR)

TIR at the top surface (block → air) occurs when $\phi$ ≥ critical angle $c$, where

$$\sin c=\frac{n_{\text{air}}}{n_{\text{block}}}=\frac{1}{1.25}=\mathrm{0.8.}$$

Thus $c = \arcsin(0.8) \approx 53.13^\circ$.

So require:

$$\varphi ={90}^{\circ }-r\ge c\Rightarrow r\le {90}^{\circ }-c\mathrm{.}$$

Compute $90^\circ - c = 90^\circ - 53.13^\circ = 36.87^\circ$. (Note: $\sin 36.87^\circ = 0.6$.)

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Step 4 — Find maximum $\theta$

At the limiting case $r = 36.87^\circ$:

$$\sin r=0.6=0.8\sin \theta \Rightarrow \sin \theta =\frac{0.6}{0.8}=0.75=\frac{3}{4}\mathrm{.}$$

So the maximum allowable $\theta$ is

$${\theta }_{\max }={\sin }^{-1}\text{\hspace{0.17em}⁣}\left(\frac{3}{4}\right)\approx {48.59}^{\circ }\mathrm{.}$$

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🖼️ Image Solution

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✅ Conclusion & Video Solution

✅ Final Answer:

$$\boxed{{\displaystyle {\theta }_{\max }={\sin }^{-1}\text{\hspace{0.17em}⁣}\left(\frac{3}{4}\right)\approx {48.6}^{\circ }\mathrm{.}}}$$

Key idea recap:

• Use Snell’s law at entry to relate $\theta$ and internal angle $r$.

• For a rectangular block the incidence angle on the horizontal top is $\phi=90^\circ-r$.

• Impose $\phi\ge$ critical angle $\arcsin(n_{\text{air}}/n_{\text{block}})$ to get the limit.