Question

When a small lamp is held 1.5 m above the surface of water in a tank, its image formed by reflection at the surface appears to coincide with the image of the bottom of the tank. (μ of water = 4/3) . The depth of the tank is:
A) 2m
B) 1.5m
C) 1m
D) 4m

Answer 1

Hint: Using Snell’s law we will find the ratio of apparent and real height, the height above the water is the real height while the height below the water after refraction is apparent height and with the refractive index of the water or the medium given, we use the formula as:
\[\dfrac{h}{h'}=\mu \]
where h is the real height, h’ is the apparent height and μ is the refractive index of water .

Complete step by step solution:
As given in the question, the lamp is held at a height of 1.5 m from the water surface or medium with a refractive index of \[\dfrac{4}{3}\].
The image is formed by the reflection at the surface appears to coincides with the image as given in the question with a height of h’
When using Snell’s law the image which is virtual in nature is formed at the bottom of the tank due to which the apparent depth and the depth at which the image is seen by the user from rarer medium are same. Hence, the formula used is:
\[\dfrac{h}{h'}=\mu \]
Placing the values in the above formula, we get the value of the apparent height h’ as:
\[\Rightarrow \dfrac{1.5}{\dfrac{4}{3}}=h'\]
\[\Rightarrow h'=2m\]

Therefore, the depth which is equal to the apparent height is given as \[h'=2m\].

Note: The apparent depth and depth of the tank or place where the medium is kept can be same under two conditions first total internal reflection or when the light passes through the medium at 90 degrees straight and the image is seen by the user at the bottom of the tank or place as the water is reflected inside on the inside surface of the water.