Question

A thin convergent glass lens $({\mu }_g=1.5)$ has a power of $+5.0\text{D}$. When this lens is immersed in a liquid of refractive index ${\mu }_l$, it acts as a divergent lens of focal length $100\text{cm}$. The value of ${\mu }_l$ is

• A. $4/3$
• B. $5/3$
• C. $5/4$
• D. $6/5$

Answer 1

The correct option is B $5/3$

Given,For convergent lens,
Refractive index of air, ${\mu }_a=1$
Refractive index of glass, ${\mu }_g=1.5$
Power of lens, $P=+5\text{D}$

So, its focal length, $$f=\frac{1}{P}$$
$\Rightarrow f=\frac{100}{5}=20\text{cm}$

Using Lens makers formula,

$\frac{1}{f}=\left(\frac{{\mu }_g-{\mu }_a}{{\mu }_a}\right)(\frac{1}{R_1}-\frac{1}{R_2})$

Substituting the values given in the problem,

$\Rightarrow \frac{1}{20}=(1.5-1)(\frac{1}{R_1}-\frac{1}{R_2}).....\left(1\right)$

Now, When convergent lens immersed in a liquid of refractive index ${\mu }_l$, convergent lens act as a diverging lens of focal length $-100\text{cm}$.

By lens maker formula,

$\frac{1}{-100}=\left(\frac{1.5-{\mu }_l}{{\mu }_l}\right)(\frac{1}{R_1}-\frac{1}{R_2}).....\left(2\right)$

Dividing $\left(1\right)$ by $\left(2\right)$ we get,

$-5=\frac{\frac{1}{2}}{(\frac{1.5}{{\mu }_l}-1)}$

$\Rightarrow 1-\frac{1.5}{{\mu }_l}=\frac{1}{10}$

$\Rightarrow \frac{9}{10}=\frac{3}{2{\mu }_l}$

$\Rightarrow {\mu }_l=\frac{5}{3}$

Hence,option (b) is the correct answer.