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

To solve this problem, we need to find the maximum value of the incident angle \( \theta \) that allows total internal reflection at the top surface of the block. The phenomenon occurs when light attempts to travel from a denser medium to a less dense medium at an angle greater than the critical angle.

The critical angle \( \theta_C \) can be determined using Snell's Law, given by:

\(n_2 \sin \theta_C = n_1 \sin 90^\circ\)

Where \( n_2 = 1.25 \) (refractive index of the block), \( n_1 = 1.0 \) (refractive index of the surrounding medium), and \(\sin 90^\circ = 1\).

Rearranging Snell's Law to find the critical angle:

\(\sin \theta_C = \frac{n_1}{n_2} = \frac{1}{1.25} = 0.8\)

So,

\(\theta_C = \sin^{-1}(0.8)\)

However, since we need the maximum value of \( \theta \) for which total internal reflection occurs, consider the geometry in the figure:

Using trigonometric relationships and geometry, we can find that for total internal reflection:

\(\theta = 90^\circ - \theta_C\)

The maximum \( \theta \) occurs when:

\(\theta = \sin^{-1}\left(\frac{3}{4}\right)\)

Thus, the correct option is:

\(\sin^{-1}(3/4)\)

Answer 2

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 \( \theta_{max} = \sin^{-1}(3/4) \).