Question

Two light beams fall on a transparent material block at point 1 and 2 with angle \( \theta_1 \) and \( \theta_2 \), respectively, as shown in the figure. After refraction, the beams intersect at point 3 which is exactly on the interface at the other end of the block. Given: the distance between 1 and 2, \( d = \frac{4}{3} \) cm and \( \theta_1 = \theta_2 = \cos^{-1} \left( \frac{n_2}{2n_1} \right) \), where \( n_2 \) is the refractive index of the block and \( n_1 \) is the refractive index of the outside medium, then the thickness of the block is …….. cm. 

Hint:

To solve refraction problems involving transparent materials, use Snell’s Law and apply trigonometric relations to find the angles and distances.

Answer 1

Using Snell’s law, the refracted angle is related to the refractive indices as: \[ \sin \theta_1 = \frac{n_2}{2 n_1} \] By applying trigonometry and using the given values for distance and angles, we can calculate the thickness of the block \( d \). Thus, the thickness of the block is \( 3 \, {cm} \).