Question

A convex spherical refracting surface of radius $R$ separates a medium having refractive index $2.5$ from a medium having refractive index $1.5$. As an object $\left(O\right)$ is moved towards the surface, from a point far away from the surface as shown in the figure, along the central axis, its image changes from real to virtual when it is at a distance $x$ from the pole $P$ of the surface. Find $x$.

• A. $x=\frac{2R}{3}$
• B. $x=\frac{3R}{2}$
• C. $x=2R$
• D. $x=R$

Answer 1

The correct option is B $x=\frac{3R}{2}$

Considering the sign convention,
$u=-u,R=+R$
${\mu }_1=1.5$ (object medium), ${\mu }_2=2.5$

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

$\Rightarrow \frac{2.5}{v}-\frac{1.5}{-u}=\frac{2.5-1.5}{R}$

$\Rightarrow \frac{2.5}{v}=-\frac{1.5}{u}+\frac{1}{R}$

$\therefore v=\frac{2.5uR}{u-1.5R}\dots \left(1\right)$

From Eq. (1), we can conclude that,
$v=\to \infty$ if $u-1.5R=0$

$\Rightarrow u=1.5R=\frac{3}{2}R$

For $u<\frac{3R}{2}$, $v$ is -ve, i.e. image is virtual.

And for $u>\frac{3R}{2}$, $v$ becomes +ve, i.e. image is real.

Hence, $x=\frac{3R}{2}$ is the answer.

Why this question? This question tests your understanding of the general case of refraction at a spherical surface. NOTE: Remember to apply the sign convention correctly.