Question

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

• A. $1.37$
• B. $3$
• C. $5.2$
• D. $2.2$

Answer 1

The correct option is A $1.37$

Formula for focal length is:

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

Here, $\frac{1}{f}=+5D,{\mu }_2=1.5,{\mu }_1=1$

Then,

$⟹+5=\left(0.5\right)(\frac{1}{R_1}-\frac{1}{R_2})-----\left(i\right)f=100cm=1m{\mu }_2=1.5$

Let ${\mu }_1=x$.

$⟹\frac{1}{1}=\left(\frac{1.5}{x-1}\right)(\frac{1}{R_1}-\frac{1}{R_2})⟹1=\left(\frac{1.5}{x-1}\right)(\frac{1}{R_1}-\frac{1}{R_2})-----\left(ii\right)$

Dividing (i) by (ii),

$⟹5=\frac{0.5}{\left(\frac{1.5}{x-1}\right)}⟹7.5x-5=0.5⟹x=1.37$