Question

A convex spherical refracting surface with radius $R$ separates a medium having refractive index $5/2$ from air. As an object is moved towards the surface from far away from the surface along the central axis, its image

• A. changes from real to virtual when it is at a distance $R$ from the surface
• B. changes from virtual to real when it is at a distance $R$ from the surface
• C. changes from real to virtual when it is at a distance $2R/3$ from the surface
• D. changes from virtual to real when it is at a distance $2R/3$ from the surface.

Answer 1

The correct option is C changes from real to virtual when it is at a distance $2R/3$ from the surface

Here object has been moved towards surface.
Considering sign convention,
$u=-u,{\mu }_2=2.5$
$R=+R,{\mu }_1=1\text{(object medium)}$

On applying,
$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$
$\Rightarrow \frac{2.5}{v}-\frac{1}{-u}=\frac{2.5-1}{R}$
$\Rightarrow \frac{2.5}{v}=\frac{1.5}{R}-\frac{1}{u}$
$\Rightarrow \frac{2.5}{v}=\frac{1.5u-R}{uR}$
$\therefore v=\frac{2.5uR}{1.5u-R}........\left(1\right)$

From Eq. (1) we can conclude,
$v=\infty \text{if}1.5u-R=0$
$\Rightarrow u=\frac{2R}{3}$

For $u<\frac{2R}{3}$,
$v$ becomes $-ve$ i.e. image will form on the left side of pole $\left(P\right)$. Thus, image will be virtual.
For $u>\frac{2R}{3}$,
$v$ will be $+ve$. Thus, real image will be formed on the right side of pole $\left(P\right)$.
$\therefore$ Transition of image form real to virtual occur at $u=\frac{2R}{3}$

Why this question? Tip : Take care of proper sign convention while using formula. A virtual image will form if, after refraction, rays appear to intersect at some point.