Question

A fish in an aquarium, $30cm$ deep in water can see a light bulb kept $50cm$ 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 $60cm$. Then the apparent distance between the two images seen by the fish is $({{\mu }_{0}}=\dfrac{4}{3})$ .
A) $140m$
B) $\dfrac{760}{3}m$
C) $\dfrac{280}{3}cm$
D) $\dfrac{380}{3}cm$

Answer 1

Hint:This problem can be solved by knowing the real depth and the apparent depth. Real depth is defined as the actual distance of an object which is placed beneath the surface of the water and apparent depth is defined as the depth of an object which is in a denser medium as seen from the rarer medium.

Complete step-by-step solution:
Refractive index in optics is dimensionless number which in turn describes how fast light can travel in that medium.by the refractive index we can determine how much path of the light ray is reflected or bent, when passing through a medium. Snell’s law describes this phenomenon.
As the wavelength varies, the refractive index also varies because of this white light when refracted split into different colors. As the refractive index increases the speed of light in any material decreases and the refractive index is also called the index of refraction. Refractive index of Ethyl Alcohol is 1.36 and the refractive index of diamond is 2.417.
Refractive index (n) is the ratio of real depth to apparent depth
 $(n)=\dfrac{real\,depth}{apparent\,depth}$
The apparent distance of fly from the wall of the aquarium for the fish is $\mu x$
Where x is the actual distance
The apparent velocity will be given by:
$\dfrac{d(\mu x)}{dt}={{({{v}_{ap}})}_{fly}}=\mu {{v}_{fly}}$
The fish observers the velocity of the fly to be $8m{{s}^{-1}}$
The apparent relative velocity will be $8m{{s}^{-1}}$
Apparent relative velocity is equal to $8m{{s}^{-1}}$
$\begin{align}
  & {{v}_{fish}}+{{({{v}_{app}})}_{fly}}=8 \\
 & \Rightarrow 3+\mu {{v}_{fly}}=8 \\
 & \Rightarrow {{v}_{fly}}=5\times \dfrac{3}{4} \\
 & \therefore {{v}_{fly}}=3.75 \\
\end{align}$
Apparent depth = $\dfrac{\text{real depth}}{\mu }$
Apparent distance of the bulb from the fish
${{D}_{1}}=50{{\mu }_{0}}+30$
Apparent distance of the image
$\Rightarrow {{D}_{2}}=50{{\mu }_{0}}+60+30$
$\Rightarrow {{D}_{1}}+{{D}_{_{2}}}=100{{\mu }_{0}}+120$
$\Rightarrow {{D}_{1}}+{{D}_{2}}=\dfrac{400}{3}+120$
$\therefore {{D}_{1}}+{{D}_{2}}=\dfrac{760}{3}m$
${{D}_{1}}+{{D}_{2}}=253.3m$
 The apparent distance between the two images is $\dfrac{760}{3}m$
So option B is correct.

Note: Real depth is more than the apparent depth and the apparent depth is always lesser than the real depth. When the real depth increases then the apparent depth also increases. Real depth is measured by submerging a perfect ruler along with it .Refractive index of water is 1.333 and the refractive index of glass is 1.003.