Question

A convex lens of glass $\left(\mu_{g}=1.45\right)$ has a focal length $f_{a}$ in air. The lens is immersed in a liquid of refractive index $\left(\mu_{1}\right)$ 1.3. The ratio of the $f_{1} / f_{a}$ is

• A. 3.9
• B. 0.23
• C. 0.43
• D. 0.39

Answer 1

Answer: (A)

Focal length of lens when placed in air
$F_{\text {air }}=\left(\mu_{\text {lens }}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\,....(i)$
Focal length of lens when placed in water
$F_{\text {air }}=\left(\frac{\mu_{\text {lens }}}{\mu_{\text {liq }}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\,......(ii)$
Dividing Eq. (i) by Eq. (ii)
$\frac{F_{\text {liq }}}{F_{\text {air }}}=\frac{(1.45-1)}{\left(\frac{1.45}{1.3}-1\right)}=\frac{0.45}{0.15} \times 1.3=3.9$