Question

A microscope is focused on a coin lying at the bottom of a beaker. The microscope is now raised up by $1\,cm$ . To what depth should the water be poured into the beaker so that coin is again in focus? (Refractive index of water is $\dfrac{4}{3}$ )
A. $1\,cm$
B. $\dfrac{4}{3}\,cm$
C. $3\,cm$
D. $4\,cm$

Answer 1

Hint: When water is poured into the beaker, the light rays coming from the coin will get refracted. So the coin lying at the bottom will appear to be raised at a certain depth. This depth is known as the apparent depth. Now, the question says that there is a shift in depth $1\,cm$ . So the focus of the microscope has been disturbed. To get the coin again in focus, water is used as a medium.

Complete step by step solution:
When light rays travel from one medium to another, it experiences bending of rays while passing obliquely through the interface due to the difference between the optical densities of the medium. This phenomenon is known as refraction. The velocity of light rays traveling in different mediums is different. The light rays traveling from rarer to denser medium bends towards the normal and the light rays traveling from denser to rarer medium will bend away from the normal.
The law of refraction is:
At the point of incidence, the incident ray, the refracted ray, and the normal to the interface all lie on the same plane.
The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant quantity. This constant term is known as the refractive index. Mathematically, it is represented as:
$\dfrac{{\sin \,i}}{{\sin \,r}} = {n_{21}}$ ……$(1)$
Where $i = $ angle of incidence
 $r = $ angle of refraction
${n_{21}} = $ the refractive index of a medium $2$ with respect to the medium $1$
The equation $(1)$ is also known as Snell’s Law.
The Refractive index is a dimensionless quantity and is invariant with time. It tells about how fast the light travels in a medium. In another way, the refractive index is defined as the ratio of the speed of light in the medium $1$ with respect to the medium $2$ . Many natural phenomena such as the formation of a rainbow, mirage, etc. are due to refraction.
Now, in the question the following parameters are given:

The apparent shift in the depth of the beaker, ${h_1} = 1\,cm$
Refractive index of water,$n = \dfrac{4}{3}$
Refractive index of air, ${n_a} \approx 1$
Let the height of water which is poured in the beaker be, ${h_2}$ .
${h_1} = \left( {1 - \dfrac{1}{n}} \right){h_2}$
$ \Rightarrow 1 = \left( {1 - \dfrac{1}{{\left( {\dfrac{4}{3}} \right)}}} \right){h_2}$
$ \Rightarrow 1 = \left( {1 - \dfrac{3}{4}} \right){h_2}$
$ \Rightarrow {h_2} = \dfrac{1}{{\left( {1 - \dfrac{3}{4}} \right)}}$
$ \Rightarrow {h_2} = \dfrac{1}{{\left( {\dfrac{1}{4}} \right)}}$
$ \therefore{h_2} = 4\,cm$
Therefore, option D is the correct answer.

Note:
Here the density is not the mass density but it is the optical density that measures the speed of light in a medium. The denser the medium, the less will be the velocity of light passing through it. Optical instruments such as a microscope, telescope, etc. use refraction phenomenon to operate as it is made up of convex lenses which are optically denser than air.