Questions: In case of total internal reflection index of refraction (nb / na) for nb in na is 1 / Sin θcrit Sin θcrit None 1 / Sin θ^2 crit

Solution Steps

Total internal reflection occurs when a light ray traveling from a medium with a higher refractive index \( n_a \) to a medium with a lower refractive index \( n_b \) hits the boundary at an angle greater than the critical angle \( \theta_{\text{crit}} \). The critical angle is the angle of incidence above which total internal reflection occurs.

The critical angle \( \theta_{\text{crit}} \) can be derived from Snell's Law, which states: \[ n_a \sin \theta_a = n_b \sin \theta_b \] For the critical angle, \( \theta_b = 90^\circ \), so \( \sin \theta_b = 1 \). Therefore, Snell's Law becomes: \[ n_a \sin \theta_{\text{crit}} = n_b \] Solving for \( \sin \theta_{\text{crit}} \), we get: \[ \sin \theta_{\text{crit}} = \frac{n_b}{n_a} \]

The question asks for the expression of the index of refraction \( \left(\frac{n_b}{n_a}\right) \) in terms of the critical angle. From the derived formula, we have: \[ \frac{n_b}{n_a} = \sin \theta_{\text{crit}} \]

Final Answer