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It is useful to design lenses of derived focal length using surface of suitable radii of curvature. Lens maker formula helps in finding various unknowns and is widely applied.

A convex lens has 30 cm focal length in air. What is its focal length is a liquid whose refractive index is $\frac{4}{3}$ ? (Given refractive index glass is $\frac{3}{2}$)

+120 cm

The correct answer is option (2) : +120 cm

$\frac{1}{f_{air}}=(\mu_g-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

$\frac{1}{f_{liquid}}=\left(\frac{\mu_g}{\mu_l}-1\right) \left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

$\frac{f_{liq}}{f_{air}}=\frac{\frac{1}{2}}{\frac{1}{8}}=4\left(\mu_g =\frac{3}{2}\text{and}\,\,  \mu_l=\frac{4}{3}\right)$

$f_{liq}=4 \times 30 = 120 \, cm$