Question

A lens of refractive index $\mu$ is put in a liquid of refractive index ${\mu }^{\prime }$. If the focal length of lens in air is $f$, then its focal length in liquid will be

• A. $\frac{-f{\mu }^{\prime }(\mu -1)}{({\mu }^{\prime }-\mu )}$
• B. $\frac{-f({\mu }^{\prime }-\mu )}{{\mu }^{\prime }(\mu -1)}$
• C. $\frac{-{\mu }^{\prime }(\mu -1)}{f({\mu }^{\prime }-\mu )}$
• D. $\frac{f{\mu }^{\prime }\mu }{(\mu -{\mu }^{\prime })}$

Answer 1

The correct option is A $\frac{-f{\mu }^{\prime }(\mu -1)}{({\mu }^{\prime }-\mu )}$

According to question.
here, $\mu$ = refractive index of lens,
$\mu$' = refractive index of liquid.
$f=$ focal length of lens in air.

Case I: When lens is in air,
From the lens maker's formula,
$\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2}),...\left(i\right)$

Case II: When lens is dipped in liquid.
$\frac{1}{f^{\prime }}=(\frac{\mu }{{\mu }^{\prime }}-1)(\frac{1}{R_1}-\frac{1}{R_2})$
$=\left(\frac{\mu -{\mu }^{\prime }}{{\mu }^{\prime }}\right)(\frac{1}{R_1}-\frac{1}{R_2})...\left(ii\right)$

From equation $\left(i\right)$ and $\left(ii\right)$, we get
$\therefore \frac{f^{\prime }}{f}=\frac{(\mu -1){\mu }^{\prime }}{(\mu -{\mu }^{\prime })}=\frac{-(\mu -1){\mu }^{\prime }}{({\mu }^{\prime }-\mu )}$
$f^{\prime }=\frac{-f{\mu }^{\prime }(\mu -1)}{({\mu }^{\prime }-\mu )}$