Question

Convex lens made up of glass $\left(\mu_g=1.5\right)$ and radius of curvature $R$ is dipped into the water. Its focal length will be (Refractive index of water = $4/3$ )

• A. $4R$
• B. $2R$
• C. $R$
• D. $\frac{R}{2}$

Answer 1

Answer: (A)

Here, $\mu_{ g }=1.5=\frac{3}{2}, \mu_{w, x}=\frac{4}{3}$
$R_{1}=R, \Rightarrow R_{2}=-R$
From, lens maker's formula
$\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$ where $\mu=\frac{\mu_{L}}{\mu_{M}}$
$\therefore \frac{1}{f_{a_{a}}}=\left(\frac{\mu_{ g }}{\mu_{a_{n}}}-1\right)\left(\frac{1}{R}-\frac{1}{(-R)}\right)$
$\Rightarrow \frac{1}{f_{a}}=\frac{1}{R}$
Also
$\frac{1}{f_{w r}}=\left(\frac{\mu_{ g }}{\mu_{u_{a}}}-1\right)\left(\frac{1}{R}-\frac{1}{(-R)}\right)$
Or $\frac{1}{f_{w}}=\frac{1}{4 R}$
Or $f_{w, x}=4 R$