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.
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To solve refraction problems involving transparent materials, use Snell’s Law and apply trigonometric relations to find the angles and distances.
According to Snell's law, the refracted angle is defined by the refractive indices as:
\[
\sin \theta_1 = \frac{n_2}{2 n_1}
\]
The thickness of the block \( d \) can be determined through trigonometric calculations using the provided distance and angle measurements.
The calculated thickness of the block is \( 3 \, {cm} \).
