The focal length of a thin lens made from a m | vedclass

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Question

(D) The Lens Maker's Formula is given by $\frac{1}{f} = (n_{rel} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$,where $n_{rel} = \frac{n_{lens}}{n

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(D) The Lens Maker's Formula is given by $\frac{1}{f} = (n_{rel} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$,where $n_{rel} = \frac{n_{lens}}{n_{medium}}$.In air $(n_a = 1)$: $\frac{1}{15} = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight) = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.Thus,$\left( \frac{1}{R_1} - \frac{1}{R_2} ight) = \frac{1}{15 	imes 0.5} = \frac{1}{7.5} \ cm^{-1}$.In liquid $(n_w = \frac{4}{3})$: $\frac{1}{f_w} = \left( \frac{1.5}{4/3} - 1 ight) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.$\frac{1}{f_w} = \left( \frac{4.5}{4} - 1 ight) \left( \frac{1}{7.5} ight) = \left( \frac{0.5}{4} ight) \left( \frac{1}{7.5} ight) = \frac{1}{8} 	imes \frac{1}{7.5} = \frac{1}{60}$.Therefore,$f_w = 60 \ cm$.

The focal length of a thin lens made from a material of refractive index $1.5$ is $15 \ cm$. When it is placed in a liquid of refractive index $\frac{4}{3}$,its focal length will be $..........\ cm$.

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