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

Two light rays initially in the same phase travel through two media of equal length $L$ having refractive indices $\mu_1$ and $\mu_2$ (where $\mu_1 > \mu_2$) as shown in the figure. If the wavelength of the light rays in air is $\lambda$,the phase difference of the emerging rays is given by:

An object is placed $20 \ cm$ in front of a $4 \ cm$ thick plane mirror. The image of the object is finally formed at $45 \ cm$ from the object itself. The refractive index of the material of the unpolished side of the mirror is (considering near normal incidence):

The lower half of a vessel of depth $2d \text{ cm}$ is filled with a liquid of refractive index $\mu_1$ and the upper half with a liquid of refractive index $\mu_2$. The apparent depth of the vessel seen perpendicularly is

$A$ container is half-filled with a liquid of refractive index $\mu$. The other half is filled with an immiscible liquid of refractive index $1.5\mu$. If the apparent depth of the container is $1.5$ times the actual depth, find $\mu$.

Difficult
View Solution

$A$ glass slab of thickness $3 \ cm$ is placed on an ink mark on a piece of paper. For a person looking at the mark from a distance of $5.0 \ cm$ above the top surface of the slab, the mark appears to be at a distance of $4.0 \ cm$ from the observer. The refractive index of the slab 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