Question

A denser medium of a refractive index $1.5$ has a concave surface of radius of curvature $12\text{cm}$. An object is situated in the denser medium at a distance of $9\text{cm}$ from the pole. Locate the image due to refraction in air.

• A. A real image at $8\text{cm}$
• B. A virtual image at $8\text{cm}$
• C. A real image at $4.8\text{cm}$
• D. A virtual image at $4.8\text{cm}$

Answer 1

The correct option is D A virtual image at $4.8\text{cm}$


As per the sign convention (direction of incident ray taken as +ve)
Given that,
Radius of curvature, $R=+12\text{cm}$
Refractive index of the denser medium, ${\mu }_1=1.5$
Distance of object, $u=-9\text{cm}$
At the spherical interface.
$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$
$\Rightarrow \frac{1}{v}-\frac{1.5}{-9}=\frac{1-1.5}{+12}$
or $\frac{1}{v}+\frac{1}{6}=\frac{-1}{24}$
or $\frac{1}{v}=\frac{-1}{24}-\frac{1}{6}=\frac{-1-4}{24}$
$\therefore v=\frac{-24}{5}\text{cm}=-4.8\text{cm}$
Thus, the image will appear to the observer at $4.8\text{cm}$ behind the pole. The image formed is virtual because image is on the same side of the object.

Why this question? Tip: Due to refraction light bends in medium, thus image will appear to form because light rays are not really intersecting.