Question

A transparent block A having refractive index $ \mu_2 = 1.25 $ is surrounded by another medium of refractive index $ \mu_1 = 1.0 $ as shown in figure. A light ray is incident on the flat face of the block with incident angle $ \theta $ as shown in figure. What is the maximum value of $ \theta $ for which light suffers total internal reflection at the top surface of the block ?

• A. \( \tan^{-1}(4/3) \)
• B. \( \tan^{-1}(3/4) \)
• C. \( \sin^{-1}(3/4) \)
• D. \( \cos^{-1}(3/4) \)

Hint:

Remember to apply Snell's law at the first interface to relate the angle of incidence \( \theta \) to the angle of refraction \( r \) inside the block. Then, use the geometry to find the angle of incidence at the second interface and apply the condition for total internal reflection, which involves the critical angle.

Answer 1

The Correct Option is C

Step 1: Apply Snell's Law at the air--block interface.
Let \( r \) be the angle of refraction inside the block. \[ \sin \theta = 1.25 \sin r = \frac{5}{4} \sin r \implies \sin r = \frac{4}{5} \sin \theta \] 
Step 2: Determine the condition for total internal reflection at the top surface.
The angle of incidence at the top surface is \( i = 90^\circ - r \). 
For total internal reflection to occur at the block-air interface, \( i \) must be greater than or equal to the critical angle \( \theta_C \), where \( \sin \theta_C = \frac{\mu_1}{\mu_2} = \frac{1}{1.25} = \frac{4}{5} \). 
So, we need \( 90^\circ - r \ge \theta_C \), which implies \( \sin(90^\circ - r) \ge \sin \theta_C \), or \( \cos r \ge \frac{4}{5} \). 
Step 3: Use a trigonometric identity to express \( \cos r \) in terms of \( \sin r \).
We know that \( \cos r = \sqrt{1 - \sin^2 r} \). Substituting the expression for \( \sin r \) from 
Step 1:
\[ \cos r = \sqrt{1 - \left(\frac{4}{5} \sin \theta\right)^2} = \sqrt{1 - \frac{16}{25} \sin^2 \theta} \] 
Step 4: Apply the condition for total internal reflection.
\[ \sqrt{1 - \frac{16}{25} \sin^2 \theta} \ge \frac{4}{5} \] Squaring both sides: \[ 1 - \frac{16}{25} \sin^2 \theta \ge \frac{16}{25} \] \[ 1 - \frac{16}{25} \ge \frac{16}{25} \sin^2 \theta \] \[ \frac{9}{25} \ge \frac{16}{25} \sin^2 \theta \] \[ 9 \ge 16 \sin^2 \theta \] \[ \sin^2 \theta \le \frac{9}{16} \] \[ |\sin \theta| \le \frac{3}{4} \] Since \( \theta \) is the angle of incidence \( (0^\circ \le \theta \le 90^\circ) \), \( \sin \theta \ge 0 \). \[ \sin \theta \le \frac{3}{4} \] The maximum value of \( \theta \) occurs when \( \sin \theta = \frac{3}{4} \), so \( \theta_{max} = \sin^{-1}(3/4) \).