Two light beams fall on a transparent material block at points 1 and 2 at angles $\theta_1$ and $\theta_2$ (measured from the surface), 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, $d = 4\sqrt3\ \text{cm}$ and $\theta_1 = \theta_2 = \cos^{-1}\left(\dfrac{n_2}{2n_1}\right)$, where the refractive index of the block $n_2$ is greater than the refractive index of the outside medium $n_1$. The thickness of the block is ______ cm.
Numerical value type. Enter your answer.
Answer: 6
The beams make angle $\theta$ with the surface, so the angle of incidence is $i = 90^\circ - \theta$ and $\sin i = \cos\theta = \dfrac{n_2}{2n_1}$.
Snell's law: $n_1\sin i = n_2\sin r \Rightarrow \sin r = \dfrac{1}{2}$, so $r = 30^\circ$.
By symmetry the two refracted beams meet directly below the midpoint of 1 and 2, a horizontal distance $d/2 = 2\sqrt3\ \text{cm}$ from each point:
$$t = \frac{d/2}{\tan30^\circ} = 2\sqrt3\times\sqrt3 = 6\ \text{cm}$$
Solution by Sreeraj P, M.Sc Physics