Question

A particle is inside a sphere of radius $15\text{cm}$ and refractive index $\frac{4}{3}$. If a dust particle is $10\text{cm}$ from the surface, where does the particle appear to an observer outside the sphere looking at it normally?

• A. $9\text{cm}$
• B. $12\text{cm}$
• C. $6\text{cm}$
• D. $14\text{cm}$

Answer 1

The correct option is A $9\text{cm}$


Take the direction of incident ray along the upward direction +ve.
The surface is convex for observer, measuring distance from $P$.
Radius of curvature, $R=-15\text{cm}$
Distance of particle from wall, $u=-10\text{cm}$
Refractive index of sphere, $\mu =\frac{4}{3}$
Now by using the formula
$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$
$(\because {\mu }_2=1,{\mu }_1=\frac{4}{3})$
$\Rightarrow \frac{1}{v}-\frac{4}{3(-10)}=\frac{1-\frac{4}{3}}{-15}$
or, $\frac{1}{v}+\frac{4}{30}=\frac{1}{45}$
$\frac{1}{v}=\frac{1}{45}-\frac{4}{30}=\frac{2-12}{90}$
$\therefore v=\frac{-90}{10}=-9\text{cm}$
Thus image will form $9\text{cm}$ below the point $P$.

Why this question? Tip: Always proceed by defining the +ve direction as per sign convention and then substitute in formula with proper sign.