Question

A concave mirror has 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 in cm from the surface of water. The value of d is close to

[Refractive index of water = 4/3]

• A. 8.8
• B. 9.0
• C. 9.00
• D. 9
• E. 8.80
• F. 8.7
• G. 8.70
• H. 8.75

Answer 1

Answer: (A), (B), (C), (D), (E), (F), (G), (H)

In the given case, $u=-5\text{cm}$

Focal length, f = $\frac{-R}{2}=\frac{-40}{2}=-20\text{cm}$
Now, using mirror formula,
$\Rightarrow v=+\frac{20}{3}cm$
For the light getting refracted at water surface, this image will acts as an object.
So, distance of object,
$d^{\prime }=5\text{cm}+\frac{20}{3}cm=\frac{35}{3}\text{cm}$
(below the surface). Let's assume final image at distance d after refraction.
$\frac{d}{d^{\prime }}=\frac{{\mu }_2}{{\mu }_1}$
$\Rightarrow d=d^{\prime }\left(\frac{{\mu }_2}{{\mu }_1}\right)=\left(\frac{35}{3}cm\right)\left(\frac{1}{\frac{4}{3}}\right)$
$$=\frac{35}{3}\times \frac{3}{4}cm=\frac{35}{4}cm$$
$=8.75\text{cm}\approx 8.8\text{cm}$

Ans. 8.75 cm

Why this question? Note: For refraction of rays at the water surface, the image formed by the mirror will act as an object.