Question

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

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

Answer 1

The correct option is C $10cm$

For refraction at first surface,
$u=-75cm,R_1=25cm$
${\mu }_1=1,{\mu }_2=\frac{4}{3}$
$\therefore \frac{{\mu }_2}{v^{\prime }}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R_1}$
$\frac{4/3}{v^{\prime }}-\frac{1}{(-7.5)}=\frac{4/3-1}{25}$
$\Rightarrow \frac{4}{3v^{\prime }}=\frac{1}{7.5}-\frac{1}{7.5}=0$
$\Rightarrow v^{\prime }=\infty$
It means the ray is parallel within the sphere.
For refraction at second surface,
$u=-\infty ,{\mu }_1=\frac{4}{3}$
${\mu }_2=1,R_2=-2.5cm$
$\therefore \frac{{\mu }_2}{v^{\prime }}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R_2}gives$
$\Rightarrow \frac{1}{v^{\prime }}-\frac{4/3}{(-\infty )}=\frac{1-\left(4/3\right)}{(-2.5)}$
$\Rightarrow v=+7.5cm$
i.e. Image is formed at distance $7.5cm$ to the right of $P_2$. Hence, distance of final image $P$ from centre $C=7.5+2.5=10cm$.