SATHEE: Optics Question 623

Optics Question 623

Question: A lens of refractive index $n$ is put in a liquid of refractive index $n^{'}$ of focal length of lens in air is $f$ , its focal length in liquid will be

[MP PET 1999]

Options:

A) $- \frac{f n^{'} (n - 1)}{n^{'} - n}$

B) $- \frac{f (n^{'} - n)}{n^{'} (n - 1)}$

C) $- \frac{n^{'} (n - 1)}{f (n^{'} - n)}$

D) $\frac{f n^{'} n}{n - n^{'}}$

Show Answer

Answer:

Correct Answer: A

Solution:

$\frac{1}{f} = (\frac{n - 1}{1}) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$ and $\frac{1}{f^{'}} = (\frac{n - n^{'}}{n^{'}}) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$∴$ $\frac{f^{'}}{f} = \frac{n - 1}{1} \times \frac{n^{'}}{n - n^{'}}$

$\Rightarrow$ $f^{'} = - \frac{f n^{'} (n - 1)}{n^{'} - n}$

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Optics Question 623

Question: A lens of refractive index n is put in a liquid of refractive index n′ of focal length of lens in air is f , its focal length in liquid will be

Options:

Answer:

Solution:

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Question: A lens of refractive index $n$ is put in a liquid of refractive index $n^{'}$ of focal length of lens in air is $f$ , its focal length in liquid will be