# A double convex lens $(\mu =1.5)$ has its focal length 10 cm. What will be its focal length if it is immersed in liquid of $(\mu_1=\large\frac{4}{3})$

$(a)\;\frac{10}{3}\;cm \\ (b)\;\frac{40}{3} \;cm \\ (c)\;40\;cm \\ (d)\;10\;cm$

focal length in air $\large\frac{1}{f_a}=(\mu_1 -1) \bigg[\large\frac{1}{r}-\frac{1}{-r} \bigg]$
$\qquad= \bigg(\large\frac{3}{2}-1 \bigg) \frac{2}{r}$
focal length in liquid $\large\frac{1}{fw} =\bigg(\large\frac{\mu_1}{\mu_2} -1\bigg) \bigg[ \large\frac{2}{r}\bigg]$
$\qquad= \bigg (\large\frac{3/2}{4/3} -1\bigg) \frac{2}{r}$
$\qquad= \bigg( \large\frac{9}{8} -1\bigg) \frac{2}{r}$
$\large\frac{f_w}{f_a} =\large\frac{(\Large\frac{3}{2}-1)}{(\Large\frac{9}{8}-1) }$
$\qquad= 4$
$\therefore f_w= f_a \times 4$
$\qquad= 10 \times 4$
$\qquad =40 \;cm$
Hence c is the correct answer.