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VITEEE 2010 Paper

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Question : 25 of 120

Marks: +1, -0

The power of a thin convex lens

({}_{a}n_{g}=1.5)

is 5.0 D. When it is placed in a liquid of refractive index

{}_{a}n_{l'}

then it behaves as a concave lens of focal length 100cm. The refractive index of the liquid

{}_{a}n_{l'}

will be

5/3
4/3
$\sqrt{3}$
5/4

Solution:

By using lens maker's formula,

\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)

\Rightarrow 5=(1.5-1)\left(\frac{2}{R}\right)

...(i)

If a lens of refractive index

\mu_g

is immersed in a liquid of refractive index

\mu_1,

then its focal length in liquid

\frac{1}{f_1}=({}_{l}\mu_g-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)

\Rightarrow -1=\left(\frac{1.5}{n}-1\right)\left(\frac{2}{R}\right)

...(ii)

Dividing, (i) by (ii)

-5=\frac{0.5n}{1.5-n}

\Rightarrow -7.5+5n=0.5n

\Rightarrow -7.5=-4.5n

\Rightarrow n=\frac{75}{45}=\frac{5}{3}

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