A thin lens made of glass (refractive index = | vedclass

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(C) Using the Lens Maker's Formula: $\frac{1}{f} = (\mu_{g} - 1) \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)$.For the lens in air $(\mu_{a} = 1)$

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(C) Using the Lens Maker's Formula: $\frac{1}{f} = (\mu_{g} - 1) \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} ight)$.For the lens in air $(\mu_{a} = 1)$:$\frac{1}{f} = (1.5 - 1) \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} ight) = 0.5 \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} ight) \dots (1)$For the lens in liquid $(\mu_{l} = 1.42)$:$\frac{1}{f_{l}} = \left( \frac{1.5}{1.42} - 1 ight) \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} ight) = \left( \frac{1.5 - 1.42}{1.42} ight) \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} ight) = \frac{0.08}{1.42} \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} ight) \dots (2)$Dividing equation $(1)$ by equation $(2)$:$\frac{f_{l}}{f} = \frac{0.5 \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} ight)}{\frac{0.08}{1.42} \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} ight)} = \frac{0.5 	imes 1.42}{0.08} = \frac{0.71}{0.08} = 8.875$.The value $8.875$ is closest to the integer $9$.

$A$ thin lens made of glass (refractive index $= 1.5$) of focal length $f = 16 \; cm$ is immersed in a liquid of refractive index $1.42$. If its focal length in liquid is $f_{l}$,then the ratio $f_{l} / f$ is closest to the integer:

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