Question

A thin walled glass sphere of radius $2.5\, cm$ is filled with water. An object $(O)$ is placed at $7.5\, cm$ from the surface of the sphere. Neglecting the effect of glass wall, at what distance the image $(I)$ of the object, measured from the centre of sphere is formed?

• A. 20 cm
• B. 15 cm
• C. 10 cm
• D. 7.5 cm

Answer 1

Answer: (C)

For refraction at first surface,
$u=-75\, cm , R_{1}=25\, cm$
$\mu_{1}=1, \mu_{2}=\frac{4}{3}$
$\therefore \frac{\mu_{2}}{v'}-\frac{\mu_{1}}{u}=\frac{\mu_{2}-\mu_{1}}{R_{1}}$
$\Rightarrow \frac{4 / 3}{v'}-\frac{1}{(-75)}=\frac{4 / 3-1}{25}$
$\Rightarrow \frac{4}{3 v'}=\frac{1}{75}-\frac{1}{75}=0$
$\Rightarrow v'=\infty$
It means the ray is parallel within the sphere.
For refraction at second surface,
$u=-\infty, \mu_{1}=\frac{4}{3}$
$\mu_{2}=1, R_{2}=-25\, cm$
$\therefore \frac{\mu_{2}}{v^{\prime}}-\frac{\mu_{1}}{u}=\frac{\mu_{2}-\mu_{1}}{R_{2}} $ gives
$\frac{1}{v^{\prime}}-\frac{4 / 3}{(-\infty)}=\frac{1-\left(\frac{4}{3}\right)}{(-25)}$
$\Rightarrow v=+7.5 cm$
i.e. Image is formed at distance $7.5\, cm$ to the right of $P_{2}$.
Hance, distance of final image $P$ from centre $C=75+25=10\, cm$