Question

A mark at the bottom of a liquid appears to rise by $0.1$ m. The depth of the liquid is $1$m. The refractive index of the liquid is –
$\left( A \right)1.33$
$\left( B \right)\dfrac{9}{{10}}$
$\left( C \right)\dfrac{{10}}{9}$
$\left( D \right)1.5$

Answer 1

Hint: In this question, we need to apply the concept of apparent depth and real depth of objects in the water. apparent depth in a medium is the depth of an object in a denser medium as seen from the rarer medium.

Complete answer:
Now using the concept of apparent depth and real depth we can proceed to the following solution.
According to the question-
Height of the appearing mark from the bottom =$0.1m$.
Apparent depth of the mark =$1 - 0.1 = 0.9m$
Here we will use the formula where the refractive index is the ratio of real depth is to the apparent depth of that substance
Now using the above-mentioned formula in the hint, we get:
Refractive index = real depth divided by the apparent depth
Now putting the known value in the above equation, we will get,
Refractive index = $\dfrac{{1m}}{{0.9m}}$= $\dfrac{{10}}{9}$
Now putting the known value in the above equation, we will get,
Refractive index = $\dfrac{{1m}}{{0.9m}}$
Refractive index =$\dfrac{{10}}{9}$
Thus, we get refractive index as
$\dfrac{10}{9}$
Here remember that the refractive index is a unitless parameter as the same unit term is canceled out.

Therefore the correct option is (C).

Note:
Refractive index has no units because we are dividing the parameter of the same unit. It is important to note the fact that this phenomenon happens only when we see any object in a denser medium from a rarer medium.