Question

A concave mirror has radius of curvature of $40cm$. It is at the bottom of a glass that has water filled up to $5cm$ (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 : (Refractive index of water $=1.33$)

• A. $8.8cm$
• B. $11.7cm$
• C. $6.7cm$
• D. $13.4cm$

Answer 1

Answer: (A)

Light incident from particle P will be reflected at mirror $u=-5cm,f=-\frac{R}{2}=-20cm$
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$v_1=+\frac{20}{3}cm$
This image will act as object for light getting refracted at water surface
So, object distance $d=5+\frac{20}{3}=\frac{35}{3}cm$ below water surface.
After refraction, final image is at
$d^{\prime }=d\left(\frac{{\mu }_2}{{\mu }_1}\right)$
$=\left(\frac{35}{3}\right)\left(\frac{1}{4/3}\right)$
$=\frac{35}{4}=8.75cm$
$\approx 8.8cm$