Question

A convex lens of focal length 10 cm and refractive index 1.5 is dipped in a liquid of refractive index 1.75. It will behave as

• A. a convex lens of focal length 10 cm
• B. a convex lens of focal length 35 cm
• C. a concave lens of focal length 10 cm
• D. a concave lens of focal length 35 cm

Answer 1

Answer: (D)

Given, focal length of convex lens, $f = 10\, cm$
$\mu_{\text{lens}} = 1.5 $ and $\mu_{\text{liquid}} = 1.75$
As, lens maker’s formula
$\frac{1}{f} = \left(\mu_{lens} -1\right) \left[\frac{1}{R_{1}} -\frac{1}{R_{2}}\right] $
$\because$ For a equi convex lens $R_{1} = R$ and $R_{2} = - R $
$ \frac{1}{f} = \left(\mu_{lens}\right)\left[\frac{1}{R} -\frac{1}{\left(-R\right)}\right] $
$\Rightarrow \frac{1}{f} = \left(\mu_{lens} -1\right) \frac{2}{R} $
$ \Rightarrow \frac{1}{10} = \left(1.5 -1\right) \frac{2}{R} $
$\Rightarrow R =10 \,cm $
If lens is dropped in liquid, then
$ \frac{1}{f'}=\left[\frac{\mu_{lens}}{\mu_{liq}} -1\right] \frac{2}{R}$
$ \Rightarrow \frac{1}{f'} = \left[\frac{1.5}{1.75} -1\right] \frac{2}{10} $
$ \Rightarrow f' = -35.71 \,cm $
Hence, the lens behaves like a concave(negative focal length) lens of focal length $35 \,cm.$