Question

A equi-convex lens is made of glass of refractive index $1.5$ and has a focal length $10\text{cm}$ in air. The lens is cut into two equal halves along a plane perpendicular to its principal axis to yield two plano-convex lenses. If one of the plano-convex lens is immersed in water the new focal length of it is $({\mu }_w=\frac{4}{3})$

• A. $40\text{cm}$
• B. $60\text{cm}$
• C. $80\text{cm}$
• D. $100\text{cm}$

Answer 1

The correct option is C $80\text{cm}$

Given,
Refractive index of lens, $\mu =1.5=\frac{3}{2}$

Refractive index of water, ${\mu }_w=\frac{4}{3}$

The relation between focal length $f$ and radius of curavature $R$ for equi-convex lens
$$\frac{1}{f}=(\mu -1)\frac{2}{R}...\left(1\right)$$ After cutting i.e two plano-convex lens. The new focal length
$$\frac{1}{f^{{}^{\prime }}}=(\mu -1)\frac{1}{R}...\left(2\right)$$
From equation $\left(1\right)$ and $\left(2\right)$

$\frac{f^{{}^{\prime }}}{f}=2$

$\Rightarrow f^{{}^{\prime }}=2f$

$\Rightarrow f^{\prime }=2\times 10\text{cm}=20\text{cm}$

When it is immersed in water, the new focal length $f_n^{{}^{\prime }}$ is
$$\frac{1}{f_n^{\prime }}=(\frac{\mu }{{\mu }_w}-1)\frac{1}{R}...\left(3\right)$$
From equation $\left(2\right)$ and $\left(3\right)$,

$\frac{f_n^{{}^{\prime }}}{f^{{}^{\prime }}}=\frac{(\mu -1)\frac{1}{R}}{(\frac{\mu }{{\mu }_w}-1)\frac{1}{R}}$

Substituting the values,

$\Rightarrow \frac{f_n^{{}^{\prime }}}{f^{{}^{\prime }}}=\frac{(\frac{3}{2}-1)}{(\frac{3}{2}\times \frac{3}{4}-1)}=4$

$\Rightarrow f_n^{{}^{\prime }}=4f^{{}^{\prime }}$

$\Rightarrow f_n^{\prime }=4\times 20$

$\Rightarrow f_n^{{}^{\prime }}=80\text{cm}$

Hence, option (c) is correct answer.