Question

A thin equiconvex glass lens of refractive index $1.5$ has power of $5D$. When the lens is immersed in a liquid of refractive index $\mu$, it acts as a divergent lens of focal length $100cm$. The value of $\mu$ of liquid is

• A. $4/3$
• B. $3/4$
• C. $15/11$
• D. $8/3$

Answer 1

The correct option is C $15/11$

Formula for focal length

$\frac{1}{f}=(\frac{{\mu }_2}{{\mu }_1}=1)(\frac{1}{R_1}-\frac{1}{R_2})$

Here, $\frac{1}{f}=50{\mu }_2=1.5{\mu }_1=1$

$5=(1.5-1)[\frac{1}{R_1}-\frac{1}{R_2}]$ ........ $\left(1\right)$

$\therefore f=100cm=1m$

${\mu }_2=1.5$ let ${\mu }_1=x$

$\therefore \frac{1}{1}=(\frac{1.5}{x}-1)(\frac{1}{R_1}-\frac{1}{R_2})$ ........ $\left(2\right)$

Dividing $\left(1\right)$ by $\left(2\right)$

$\frac{5}{1}=\frac{0.5}{\frac{1.5}{x}-1}$

$\therefore \frac{1.5}{x}-1=\frac{0.5}{5}$

$\therefore \frac{1.5}{x}=\frac{11}{10}$

$\therefore x=\frac{15}{11}\therefore {\mu }_1=\frac{15}{11}=1.37$