Question

Sunlight travels from medium A (refractive index = 1.5) to medium B (refractive index = 1.033). The angle of incidence beyond which the refracted ray will be in medium A without travelling to medium B is ________ ° (Answer in decimal degrees and rounded off to two decimal values).

Hint:

This is a total-internal-reflection question; the critical angle from the denser medium A into the rarer medium B satisfies sin(critical angle) equal to n_B divided by n_A.

Answer 1

Correct Answer: 43.5

Step 1: Recognise this as a critical-angle problem.
Light is travelling from medium A (denser, refractive index \(n_A = 1.5\)) into medium B (rarer, refractive index \(n_B = 1.033\)). Since \(n_A > n_B\), there exists a critical angle of incidence beyond which the ray undergoes total internal reflection and never enters medium B; it stays travelling within medium A.

Step 2: Apply Snell's law at the critical condition.
Snell's law states \( n_A \sin\theta_i = n_B \sin\theta_r \). Total internal reflection begins exactly when the refracted ray grazes the interface, i.e. \(\theta_r = 90^\circ\), so \(\sin\theta_r = 1\). The angle of incidence at this condition is the critical angle \(\theta_c\): \[ n_A \sin\theta_c = n_B(1) \implies \sin\theta_c = \frac{n_B}{n_A} \]

Step 3: Substitute the given refractive indices.
\[ \sin\theta_c = \frac{1.033}{1.5} = 0.68867 \]

Step 4: Take the inverse sine.
\[ \theta_c = \sin^{-1}(0.68867) \approx 43.52^\circ \]

Step 5: Interpret the result.
For any angle of incidence greater than \(43.52^\circ\), the refracted ray cannot cross into medium B and the ray is totally internally reflected, remaining within medium A.

\[ \boxed{\theta_c \approx 43.52^\circ} \]