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 ascertain the block's thickness, Snell's Law is applied at entry points 1 and 2, with incident angles \( \theta_1 \) and \( \theta_2 \). Snell's Law is given by:

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

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

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

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

Rearranging Snell's Law for the refracted angle yields:

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

Considering the geometry with the refraction angle established:

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

The thickness \( t \) is derived from the geometry of the refracted light path:

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

The tangent of the refracted angle is calculated as:

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

The thickness \( t \) is then computed as:

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

Consequently, the block's thickness is determined to be 6 cm.