Question

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

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

Answer 1

Answer: (A)

$8.8 \,cm$

Answer 2

Light incident from particle P will be reflected at mirror $u = -5cm , f=- \frac{R}{2} = - 20 cm$
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
$v_{1} = + \frac{20}{3} cm$
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
$d' = d \left(\frac{\mu_{2}}{\mu_{1}}\right)$
$= \left(\frac{35}{3}\right) \left(\frac{1}{4/3}\right)$
$= \frac{35}{4} = 8.75 cm$
$\approx8.8 cm$