Question

As shown in the figure, region $BCDEF$ and $ABFG$ are of refractive index $2.0$ and $1.5$ respectively. A particle $O$ is kept at the mid-point of $DH$. Image of the object as seen by the eye is at a distance of :

Answer 1

Here the object is placed in medium $(\mu =2),$
Thus, ${\mu }_1=2$
Observer is in medium, ${\mu }_2=1$

Sign convention: Direction of incident ray is taken as +ve.
(ray emerging from $O$ and moving toward $D$ is positive.)

By using the equation :

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$

Here, $u=\frac{-20}{2}=-10\text{cm}$

$R=-10\text{cm}$

$\Rightarrow \frac{1}{v}-\frac{2}{(-10)}=\frac{1-2}{-10}$

or, $\frac{1}{v}+\frac{2}{10}=\frac{1}{10}$

or, $\frac{1}{v}=\frac{1}{10}-\frac{2}{10}=\frac{-1}{10}$

$\therefore v=-10\text{cm}$

Thus, image is formed at $10\text{cm}$ from point $D$.

Why this question? Bottom line: The observer (in medium ${\mu }_2=1$) will see the image of object (in medium ${\mu }_1=2)$, due to refraction from spherical surface $CDE$.