Question

The interface AB between the two media A and B is shown in the figure. In the denser medium A, the incident ray PQ makes an angle of \( 30^\circ \) with the horizontal. The refracted ray is parallel to the interface. The refractive index of medium B with respect to medium A is:

Answer 1

 

For the given case, using Snell’s law: \[ n_A \sin \theta_A = n_B \sin \theta_B \] Here, since the refracted ray is parallel to the interface, \( \theta_B = 90^\circ \), and hence: \[ n_A \sin 30^\circ = n_B \sin 90^\circ \] \[ n_A \times \frac{1}{2} = n_B \times 1 \] Thus, the refractive index of medium B is: \[ n_B = \frac{n_A}{2} \] Since the refractive index of the denser medium (A) is \( \sqrt{\frac{4}{3}} \), we have: \[ n_B = \frac{\sqrt{\frac{4}{3}}}{2} = \frac{2}{\sqrt{3}} \] Thus, the correct answer is: \[ \text{(D) } \frac{2}{\sqrt{3}} \]