Question

A swimming pool appears to be raised by $6\,{\text{m}}$ when viewed normally. What is the refractive index of water if the actual depth of the swimming pool is $24\,{\text{m}}$?
A. 1.44
B. 1.33
C. 1.56
D. 1.76

Answer 1

Calculate the apparent depth of the swimming pool using the formula relating real depth and raised depth. Then use the formula of refractive index of a medium which is in terms of the actual depth and apparent depth.

Formula used:
The apparent depth of an object in a medium is equal to the subtraction of the real (actual) depth of the object and the raised depth of the object when viewed normally.
${\text{Apparent depth}} = \left( {{\text{Real depth}}} \right) - \left( {{\text{Raised depth}}} \right)$ …… (1)
The refractive index $\mu$ of a medium is given by
$\mu = \dfrac{{{\text{Real depth}}}}{{{\text{Apparent depth}}}}$ …… (2)

Complete step by step answer:
The actual depth of an object is the depth of the object from the surface measured with a real ruler or any other measuring equipment.
The apparent depth of an object in a denser medium is the depth of the object when seen from a rarer medium.
The value of the apparent depth of an object is always less than the real depth of the object.
The actual depth of the swimming pool is $24\,{\text{m}}$ and it appears to be raised by $6\,{\text{m}}$ when viewed normally.
Calculate the apparent depth of the swimming pool.
Substitute $24\,{\text{m}}$ for ${\text{Real depth}}$ and $6\,{\text{m}}$ for ${\text{Raised depth}}$ in equation (1).
${\text{Apparent depth}} = \left( {24\,{\text{m}}} \right) - \left( {6\,{\text{m}}} \right)$
$\Rightarrow {\text{Apparent depth}} = 18\,{\text{m}}$

Hence, the apparent depth of the swimming pool when viewed normally is $18\,{\text{m}}$.
Now, calculate the refractive index of the water.
Substitute $24\,{\text{m}}$ for ${\text{Real depth}}$ and $18\,{\text{m}}$ for ${\text{Apparent depth}}$ in equation (2).
$\mu = \dfrac{{24\,{\text{m}}}}{{18\,{\text{m}}}}$
$\Rightarrow \mu = 1.33$
Therefore, the refractive index of the water is $1.33$.

So, the correct answer is “Option B”.

Note:
The apparent depth of the swimming pool is less than the actual depth because the light undergoes refraction at the water surface boundary (air-water interface).