Question

A concave mirror has a radius of curvature of $40\text{cm}$. It is at the bottom of a glass that has water filled up to $5\text{cm}$ (see figure). If a small particle is floating on the surface of water, its image as seen, from directly above the glass, is at a distance $d$ from the surface of water. The value of $d$ is close to :
$(\text{Refractive index of water}=1.33)$

• A. $11.7\text{cm}$
• B. $8.8\text{cm}$
• C. $13.4\text{cm}$
• D. $6.7\text{cm}$

Answer 1

The correct option is B $8.8\text{cm}$

$\text{Given that,}R=40\text{cm}\Rightarrow f=20\text{cm}$

If $v$ is the distance of the image formed by the mirror, then, using mirror formula,

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

$\Rightarrow \frac{1}{v}+\frac{1}{-5}=\frac{1}{-20}$

$\Rightarrow v=\frac{20}{3}\text{cm}$

Distance of this image from water surface

$D=\frac{20}{3}+5=\frac{35}{3}\text{cm}$

Viewing from above the glass, Apparent depth due to refraction is given by,

$d=\frac{D}{\mu }$

$\Rightarrow d=\frac{\left(\frac{35}{3}\right)}{1.33}=8.8\text{cm}$

Hence, option $\left(C\right)$ is correct.