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)
$\frac{F_{\text{liq }}}{F_{\text{air }}}=\frac{(1.45-1)}{\left(\frac{1.45}{1.3}-1\right)}=\frac{0.45}{0.15}\times 1.3=3.9$