A thin convex lens of refractive index $1.5$ has $20 c m$ focal length in air. If the lens is completely immersed in a liquid refractive index $1.6$ , its focal length will be :

- 160 c m

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- 100 c m

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+ 10 c m

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+ 100 c m

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Solution

The correct option is C $- 160 c m$
According to lens maker's formula,

$\frac{1}{f_{1}} = [\frac{(n_{2} - n_{1})}{n_{1}}] [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

Here, $f_{1}$ = 20 cm, $n_{1}$ = 1(air) and $n_{2}$ = 1.5

$\frac{1}{20} = [\frac{(1.5 - 1)}{1}] [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

$\frac{1}{20} = (0.5) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$ ...............(1)

Also,

$\frac{1}{f_{2}} = [\frac{(n_{2} - n_{1})}{n_{1}}] [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

Here, $n_{1}$ = 1.6(liquid) and $n_{2}$ = 1.5

$\frac{1}{f_{2}} = [\frac{(1.5 - 1.6)}{1.6}] [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

$\frac{1}{f_{2}} = (- 0.0625) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$ ...............(2)

Dividing equation (1) by (2), we get

$\frac{f_{2}}{20} = \frac{0.5}{(- 0.0625)}$

$f_{2} = \frac{10}{(- 0.0625)}$

$f_{2} = - 160 c m$

Hence, correct answer is option A.

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A thin convex lens of refractive index 1.5 has 20cm focal length in air. If the lens is completely immersed in a liquid refractive index 1.6, its focal length will be :

A thin convex lens of refractive index $1.5$ has $20 c m$ focal length in air. If the lens is completely immersed in a liquid refractive index $1.6$ , its focal length will be :