Using the lens maker’s formula:
\[\frac{1}{f} = \left( \frac{\mu_{\text{lens}}}{\mu_{\text{air}}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)\]
Given:
\[f = 20 \, \text{cm}, \quad R_1 = 15 \, \text{cm}, \quad R_2 = -30 \, \text{cm}\]
Substitute into the formula:
\[\frac{1}{20} = (\mu - 1) \left( \frac{1}{15} - \frac{1}{-30} \right)\]
Simplify the expression:
\[\frac{1}{20} = (\mu - 1) \left( \frac{3}{30} \right)\]
\[\Rightarrow \mu - 1 = \frac{1}{2}\]
\[\Rightarrow \mu = 1 + \frac{1}{2} = \frac{3}{2} = 1.5\]
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.