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
$\frac{1}{f_{wr}}=\left(\frac{{\mu }_g}{{\mu }_{u_a}}-1\right)\left(\frac{1}{R}-\frac{1}{(-R)}\right)$
Or $\frac{1}{f_w}=\frac{1}{4R}$
Or $f_{w,x}=4R$