Question

A fish in an aquarium, $30\text{cm}$ deep in water $(\mu =4/3)$ can see a light bulb kept $50\text{cm}$ above the surface of water. The fish can also see the image of this bulb in the reflecting bottom surface of the aquarium. Total depth of water is $60\text{cm}$. Then, the apparent distance between the two images seen by the fish is

• A. $\frac{730}{3}\text{cm}$
• B. $\frac{760}{3}\text{cm}$
• C. $\frac{710}{3}\text{cm}$
• D. $\frac{770}{3}\text{cm}$

Answer 1

The correct option is B $\frac{760}{3}\text{cm}$

The fish (observer) is in denser medium, hence, bulb will appear at a farther distance for fish.


Apparent distance of bulb for fish,
$d_1=d+\mu h=30+\frac{4}{3}\times 50=\frac{290}{3}\text{cm}$ ..........$\left(1\right)$
The bottom surface of aquarium is reflecting, hence, for mirror the object distance will be,
$u=$ height of water $+$ apparent height of bulb
$u=60+\frac{4}{3}\times 50=\frac{380}{3}\text{cm}$
For a plane mirror,
$|u|=|v|$
Thus, the distance of the image of the bulb from the mirror,
$v=\frac{380}{3}\text{cm}$ ..........$\left(2\right)$
Now, the apparent distance between the two images,
$s=\frac{290}{3}+30+\frac{380}{3}=\frac{760}{3}\text{cm}$
$[$from $\left(1\right)$ and $\left(2\right)]$

Why this question? It challenges you to apply your understanding of apparent depth and image formation by plane mirror.