Question

The focal length of a biconvex lens of refractive index 3/2 is 18 cm .Its focal length in meters when immersed in a liquid of refractive index 4/3 is

• A. 160
• B. 16
• C. 32
• D. 0.5

Answer 1

The correct option is C 32

$\begin{array}{c}{f}_1=18m\\ {\mu }_g=3/2\\ f_2=?\\ {\mu }_L=\frac{4}{3}\end{array}$

$\begin{array}{c}\text{using Lens makers}\\ \text{formula and taking}\\ \text{ratios.}\end{array}$

$\begin{array}{c}{f}_1=\frac{1}{({\mu }_1-1)}\\ f_2=\frac{1}{({\mu }_2-1)}-\left(i\right)\end{array}$

$\text{Dividing (i) by (ii)}$

$\begin{array}{cc}& \frac{f_1}{f_2}=\frac{({\mu }_2-1)}{({\mu }_1-1)}\\ & \frac{18}{f_2}=\left(\frac{\frac{\frac{3}{2}}{\frac{4}{3}}-1}{\frac{3}{2}-1}\right)\\ & \frac{18}{t}=\frac{\frac{3}{2}-\frac{4}{3}}{\frac{\frac{4}{3}}{\frac{3-2}{2}}}\\ & f_2=32cm\end{array}$