Question

If the refractive index from air to glass is $\frac{3}{2}$ and that from air to water is $\frac{4}{3}$, then the ratio of focal lengths of a glass lens in water and in air is

• A. $1:2$
• B. $2:1$
• C. $1:4$
• D. $4:1$

Answer 1

Answer: (D)

$n_w=\frac{4}{3}{n}_g=\frac{3}{2}$
$f_a\left(n_g-1\right)=f_w\left(\frac{n_g}{n_w}-1\right)$
$\frac{f_w}{f_a}=\frac{n_g-1}{\frac{n_g}{n_w}-1}=\frac{\frac{3}{2}-1}{\frac{\frac{3}{2}}{\frac{4}{3}}-1}$
$=\frac{\frac{3-2}{2}}{\frac{9}{8}-1}=\frac{\frac{1}{2}}{\frac{9-8}{8}}=\frac{\frac{1}{2}}{\frac{1}{8}}=\frac{8}{2}$
$\therefore \frac{f_w}{f_a}=\frac{4}{1}$