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.
Show Hint
To solve refraction problems involving transparent materials, use Snell’s Law and apply trigonometric relations to find the angles and distances.
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 \( 3 \, {cm} \).
