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{fn’(n-1)}{n’-n} $
B) $ -\frac{f(n’-n)}{n’(n-1)} $
C) $ -\frac{n’(n-1)}{f(n’-n)} $
D) $ \frac{fn’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}} ) $
$ \therefore $ $ \frac{f’}{f}=\frac{n-1}{1}\times \frac{n’}{n-n’} $
$ \Rightarrow $ $ f’=-\frac{fn’(n-1)}{n’-n} $
