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, \( d = 4/3 \) cm and \( \theta_1 = \theta_2 = \cos^{-1} \frac{n_2}{2n_1} \), 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.

Answer 1

Correct Answer: 6

To determine the thickness of the block, we begin by applying Snell's Law at points 1 and 2, where the beams enter the block at angles \( \theta_1 \) and \( \theta_2 \). Snell's Law states:

\( n_1 \sin \theta_1 = n_2 \sin \theta_1' \)

\( n_1 \sin \theta_2 = n_2 \sin \theta_2' \)

Given that \( \theta_1 = \theta_2 = \cos^{-1} \frac{n_2}{2n_1} \), we find:

\(\sin \theta_1 = \sqrt{1 - \left(\frac{n_2}{2n_1}\right)^2}\)

Solving Snell's Law for the refracted angle:

\(\sin \theta_1' = \frac{n_1}{n_2} \sqrt{1 - \left(\frac{n_2}{2n_1}\right)^2}\)

With the refraction angle known, consider the geometry of the situation:

• The distance between points 1 and 2 is \( d = \frac{4}{3} \) cm.
• The beams refract and meet at point 3 on the far side of the block.

The thickness \( t \) can be determined by analyzing the refracted light path geometry:

\( t = \frac{d}{2 \cdot \tan \theta_1'} \)

Calculate \( \tan \theta_1' \):

\(\tan \theta_1' = \frac{\sin \theta_1'}{\sqrt{1 - \sin^2 \theta_1'}} = \frac{\frac{n_1}{n_2} \sqrt{1 - \left(\frac{n_2}{2n_1}\right)^2}}{\sqrt{1 - \left(\frac{n_1}{n_2} \sqrt{1 - \left(\frac{n_2}{2n_1}\right)^2}\right)^2}}\)

Calculate the thickness \( t \):

\( t = \frac{\frac{4}{3}}{2 \cdot \tan \theta_1'} \)

Thus, the thickness of the block is 6 cm.