Question

A converging lens of refractive index 1.5 has a power of 10D . When it is completely immersed in a liquid , it behaves as diverging lens of focal length 50 cm . Find the refractive index of the liquid.

Answer 1

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Answer 2

$n=1.67$

Explanation:

Given:

refractive index of a converging lens, $n_{ag}=1.5$

power of the lens, $P=10 D$

focal length of the lens after immersing it in a liquid, $f'=-50 \cm$

Now the focal length of the lens in the air:

$f=\frac{1}{P}$

$f=\frac{1}{10}$

$f=0.1\ m=10\ cm$

using lens maker's formula

$\frac{1}{f} =(\frac{n_{ag}}{n_a}-1)(\frac{1}{R_1}-\frac{1}{R_2})$

where:

$n_a=$ refractive index of air

$R_1\ \&\ R_2$ are the radii of curvatures of the lens.

$\frac{1}{10}=(\frac{1.5}{1}-1)(\frac{1}{R_1}-\frac{1}{R_2})$

$(\frac{1}{R_1}-\frac{1}{R_2})=\frac{1}{5}$

When the lens is immersed in the liquid:

$\frac{1}{f'} =(\frac{n_{ag}}{n}-1)(\frac{1}{R_1}-\frac{1}{R_2})$

where:

n = refractive index of the liquid

$-\frac{1}{50} =(\frac{1.5}{n}-1)\times \frac{1}{5}$

$n=1.67$

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TOPIC: radius of curvature

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