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The focal length of lens of refractive index 25 in air is 30 cm. If it is immersed in a liquid of refractive index $\frac{5}{3}$ , then its focal length will be

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150 cm

30 cm

120 cm

60 cm

answer is A.

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Detailed Solution

$μ_{LG} = \frac{μ_{ag}}{μ_{aL}} = \frac{2 .5}{8} \times 3 = 1 .5$

$\frac{1}{f} = (1 .5 - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) \frac{1}{30} = 1 .5 (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$\frac{1}{f} = 0 .5 \times \frac{1}{45} = \frac{1}{150} cm \frac{1}{R_{1}} - \frac{1}{R_{2}} = \frac{1}{45}$

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