A thin convex lens made from crown glass ($\mu =\frac{3}{2}$) has focal length f. When it is measured in two different liquids having refractive indices $\frac{4}{3}$ and $\frac{5}{3}$, it has the focal lengths  $f_1$and  $f_2$ respectively. The correct relation between the focal lengths is

Given  $\mu =\frac{3}{2}$(Crown glass) and focal length =f

Focal length = $f_1$ when lens is placed in liquid of refractive index ,${\mu }_1=\frac{4}{3}$ Focal length = $f_2$ when lens is placed in liquid of refractive index ${\mu }_2=\frac{5}{3}$

 Using lens maker’s formula

$\frac{1}{f_1}=\left(\frac{\mu }{{\mu }_1}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

$\frac{1}{f_1}=\left(\frac{3/2}{4/3}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$=\frac{1}{8}\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

Similarly , $\frac{1}{f_2}=\left(\frac{\mu }{{\mu }_2}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

$\frac{1}{f_2}=\left(\frac{3/2}{5/3}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$\Rightarrow \frac{-1}{10}\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

And $\frac{1}{f}=\left(\frac{3}{2}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ $\Rightarrow \frac{1}{2}\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

Hence, $f_1=4fand{f}_2=-5f$