Question

A light ray passes from mineral A to mineral B having refractive indices of 1.750 and 1.430, respectively. The limiting value of angle of incidence above which the light ray undergoes total internal reflection is ______________ degree (rounded off to one decimal place).

Hint:

Total internal reflection occurs when the angle of incidence exceeds the critical angle, calculated using the refractive indices of the two media.

Answer 1

Correct Answer: 54.5

Step 1: Use Snell's Law to Calculate the Critical Angle.
Snell's Law gives the relationship between the angle of incidence (\(\theta_i\)) and the angle of refraction (\(\theta_r\)): \[ n_1 \sin(\theta_i) = n_2 \sin(\theta_r). \] For total internal reflection, the angle of refraction is 90°. Thus, Snell's Law simplifies to: \[ \sin(\theta_c) = \frac{n_B}{n_A}, \] where \(n_A\) is the refractive index of mineral A, and \(n_B\) is the refractive index of mineral B. Substituting the values: \[ \sin(\theta_c) = \frac{1.430}{1.750} = 0.817. \] Taking the inverse sine: \[ \theta_c = \sin^{-1}(0.817) = 55.2°. \] Step 2: Conclusion.
The limiting angle of incidence is 55.2°, above which total internal reflection occurs.