A thin convergent glass lens ($μ_g= 1.5$) has a power of $+5.0 D$. When this lens is immersed in a liquid of refractive index $μ_l$ it acts as a divergent lens of focal length 100 cm. The value of $μ_l$ must be
Answer & explanation
Correct answer: option 2
$\text{Power of a lens } P = \frac{1}{f} = (\frac{\mu_g}{\mu_a} -1)(\frac{1}{R_1} - \frac{1}{R_2} = 5D$
$\text{when this is immerged in a liquid its focal length become -100 cm}$
$ P'= (\frac{\mu_g}{\mu_l} -1)(\frac{1}{R_1} - \frac{1}{R_2} = -1D$
$\text{Dividing these two equations }$
$\frac{P}{P'} = \frac{\frac{\mu_g}{\mu_a} -1}{\frac{\mu_g}{\mu_l} -1} = -5$
$\Rightarrow \mu = \frac{5}{3}$