Question

For the arrangement of medium shown in figure, find position of second image $I_2$ formed after two successive refractions from two plane surfaces AB and CD.

• A. 25/3 cm from CD towards right
• B. 40/3 cm from CD towards left
• C. 100/3 cm from CD towards left
• D. 50/3 cm from CD towards right

Answer 1

The correct option is C 100/3 cm from CD towards left

For refraction at plane surface AB, incident ray is travelling from medium ${\mu }_1=1$ to medium ${\mu }_2=1.5$
$\Rightarrow \frac{{\mu }_1}{{\mu }_2}=\frac{u}{v}$


or, $\frac{1}{1.5}=\frac{u}{v}$
$(\because u=EO=10cm)$
$\Rightarrow v=15cm$
or $E{I}_1=15cm$
For refraction at plane surface/interface CD, image I, will behave as object for observer just outside CD.
$\Rightarrow u^{\prime }=15+10=25cm$
$\because$ Incident ray travelling from medium ${\mu }_2=1.5$ to ${\mu }_3=2$
$[\Rightarrow \frac{{\mu }_2}{{\mu }_3}=\frac{u^{\prime }}{v^{\prime }}$
or, $\frac{1.5}{2}=\frac{25}{v^{\prime }}$
$\therefore v^{\prime }=\frac{50}{1.5}=\frac{100}{3}cm$ or $E{I}_2=\frac{100}{3}cm$
Thus final image after two refractions is formed at distance of $\frac{100}{3}cm$ from CD towards left, because ${\mu }_3>{\mu }_2>{\mu }_1$, hence object will appear at a farther distance from 'O' towards left.

Why this question? It challenges you to apply fundamentals of apparent distance at both refracting surfaces.