Obtain the relation between the real depth and apparent depth of the bottom of a tank filled with water when observed from air.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) As shown in the figure,the bottom of the tank is at a real depth $h_{2}$ in a denser medium (water) of refractive index $n_{2} = n$.
When the bottom is viewed normally,it is observed at $O^{\prime}$ instead of $O$ as shown in figure $(a)$. When it is viewed at some angle from the normal,it is observed at $O^{\prime}$ as shown in figure $(b)$. The real depth is $h_{2}$ and the apparent depth is $h_{1}$.
The relation between refractive indices and depths is given by:
$\frac{\text{Refractive index of air } (n_{1})}{\text{Refractive index of denser medium } (n_{2})} = \frac{\text{Apparent depth } (h_{1})}{\text{Real depth } (h_{2})}$
Since for air $n_{1} = 1$ and for the denser medium $n_{2} = n$:
$\frac{1}{n} = \frac{h_{1}}{h_{2}}$
Therefore,the relation is:
$n = \frac{h_{2}}{h_{1}} = \frac{\text{Real depth}}{\text{Apparent depth}}$
Thus,the apparent depth is:
$h_{1} = \frac{h_{2}}{n} = \frac{\text{Real depth}}{\text{Refractive index of denser medium}}$

Explore More

Similar Questions

Three immiscible liquids of densities $d_1 > d_2 > d_3$ and refractive indices $\mu_1 > \mu_2 > \mu_3$ are put in a beaker. The height of each liquid column is $\frac{h}{3}$. $A$ dot is made at the bottom of the beaker. For near normal vision,find the apparent depth of the dot.

An empty tank has a concave mirror at its bottom. When sunlight falls normally on the mirror, it is focused at a height of $32 \,cm$ from the mirror. If the tank is filled with water up to a height of $20 \,cm$, then the sunlight focuses at (refractive index of water $= 4/3$):

$A$ flat glass slab of thickness $6\, cm$ and refractive index $1.5$ is placed in front of a plane mirror. An observer is standing behind the glass slab and looking at the mirror. The actual distance of the observer from the mirror is $50\, cm$. The distance of his image from himself,as seen by the observer is .......$cm$.

$A$ glass slab of thickness $20\, cm$ and refractive index $1.5$ is placed in front of a plane mirror. An object is placed at a distance of $40\, cm$ from the mirror in air. The position of the final image with respect to the mirror will be at a distance of:

Difficult
View Solution

$A$ ray of light is incident on a surface of a glass slab at an angle of $45^{\circ}$. If the lateral shift produced per unit thickness is $\frac{1}{\sqrt{3}}$,the angle of refraction produced is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo