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:
\[\frac{1}{20} = (\mu - 1) \left( \frac{3}{30} \right)\]
\[\Rightarrow \mu - 1 = \frac{1}{2}\]
\[\Rightarrow \mu = 1 + \frac{1}{2} = \frac{3}{2} = 1.5\]
Choose the correct set of reagents for the following conversion:
A bead of mass \( m \) slides without friction on the wall of a vertical circular hoop of radius \( R \) as shown in figure. The bead moves under the combined action of gravity and a massless spring \( k \) attached to the bottom of the hoop. The equilibrium length of the spring is \( R \). If the bead is released from the top of the hoop with (negligible) zero initial speed, the velocity of the bead, when the length of spring becomes \( R \), would be (spring constant is \( k \), \( g \) is acceleration due to gravity):