A lens of refractive index n is put in a liquid of refractive index ( n prime), if focal length of lens in air is ' f ', its focal length in liquid will be:

Question

A lens of refractive index n is put in a liquid of refractive index ( n prime), if focal length of lens in air is ' f ', its focal length in liquid will be: (A) (-f n'(n-1)/n'-n) (B) (-f(n'-n)/n'(n-1)) (C) (-n'(n-1)/f(n'-n)) (D) (f n' n/n-n')

Tardigrade Admin · Accepted Answer

(1/f)=((n/1)-1)((1/R1)-(1/R2)) (1/fl)-((n/n')-1)((1/R1)-(1/R2)) (fl/f)=((n-1) n'/n-n') or fl=(-f n'(n-1)/n'-n)

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

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

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

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

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

$\frac{1}{f} = (\frac{n}{1} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$
$\frac{1}{f _{l}} - (\frac{n}{n ^{'}} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$
$\frac{f _{l}}{f} = \frac{( n - 1 ) n ^{'}}{n - n ^{'}}$ or $f_{l} = \frac{- f n ^{'} ( n - 1 )}{n ^{'} - n}$

Q. A lens of refractive index n is put in a liquid of refractive index (n′), if focal length of lens in air is ' f ', its focal length in liquid will be:

n′−n−fn′(n−1)​

n′(n−1)−f(n′−n)​

f(n′−n)−n′(n−1)​

n−n′fn′n​