Question

A rectangular glass slab ABCD of refractive index \(n_1\) is immersed in water of refractive index \(n_2\) \((n_1>n_2)\). A ray of light is incident at the surface AB of the slab as shown. The maximum value of the angle of incidence \(\alpha_{\max}\) such that the ray comes out only from the another surface CD is given by

• A. \(\sin^{-1}\!\left[\dfrac{n_1}{n_2}\cos\!\left(\sin^{-1}\!\dfrac{n_2}{n_1}\right)\right]\)
• B. \(\sin^{-1}\!\left[n_1\cos\!\left(\sin^{-1}\!\dfrac{1}{2}\right)\right]\)
• C. \(\sin^{-1}\!\dfrac{n_1}{n_2}\)
• D. \(\sin^{-1}\!\dfrac{n_2}{n_1}\)

Hint:

When the refracted ray must undergo TIR at the next face, use the condition \(r_1 + r_2 = 90°\) and apply Snell's law at both surfaces systematically.

Answer 1

The Correct Option is A

Step 1: Underlying Concept:
For the ray to exit only from CD, it must undergo total internal reflection at the AD face. The critical angle \(\theta_C\) at the AD face satisfies \(\sin\theta_C = \dfrac{n_2}{n_1}\). Step 2: Explanation:
From geometry, \(r_1 + r_2 = 90°\), so \((r_1)_{\max} = 90° - \theta_C\). Applying Snell's law at face AB: \[ \frac{n_1}{n_2} = \frac{\sin\alpha_{\max}}{\sin(90°-\theta_C)} = \frac{\sin\alpha_{\max}}{\cos\theta_C} \] \[ \sin\alpha_{\max} = \frac{n_1}{n_2}\cos\theta_C = \frac{n_1}{n_2}\cos\!\left(\sin^{-1}\frac{n_2}{n_1}\right) \] Step 3: Conclusion:
\(\alpha_{\max} = \sin^{-1}\!\left[\dfrac{n_1}{n_2}\cos\!\left(\sin^{-1}\!\dfrac{n_2}{n_1}\right)\right]\)