A convergent lens made of crown glass (refractive index 1.5) has focal length 20cm in air. If it is immersed in a liquid of refractive index 1.60, its focal length will be:

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A thin converging lens is made up of a glass of refractive index 1.5. It acts like a concave lens of focal length 50 cm when immersed in a liquid of refractive index (158) The focal length of the converging lens in air is (in m)

The correct option is B 0.201f=(μ−1)(1R1−1R2)f1f2=(μ2−1)(μ1−1)50cmf=(1.5−1)(1−45)500.5×0.2=ff=0.2 m

A thin converging lens is made up of a glass of refractive index $1.5$ . It acts like a concave lens of focal length $50$ cm when immersed in a liquid of refractive index $(\frac{15}{8})$ The focal length of the converging lens in air is (in m)

The correct option is B $0.20$
$\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$\frac{f_{1}}{f_{2}} = \frac{(μ_{2} - 1)}{(μ_{1} - 1)}$

$\frac{50 c m}{f} = \frac{(1.5 - 1)}{(1 - \frac{4}{5})}$

$\frac{50}{0.5} \times 0.2 = f$

$f = 0.2 m$