Real and Apparent Depth - Important Concept with Solved Examples for JEE

Suppose we consider an empty beaker of certain height and we place an object at the base of the beaker. Now, the depth measured is the real depth at which the object is placed. If we put some liquid in the glass, then the measured depth would be the apparent depth. The change in the depth without the change in the position of the object is due to the phenomena of refraction. In this article, we will be understanding this concept.

What Is Refraction and Refractive Index?

Refraction is a phenomenon in which there is bending of light when it propagates from one medium to the other. From Snell’s law, we know that

1. The incident ray, refracted ray and normal to the interface at the point of incidence lie in the same plane.

2. The ratio of sine of angle of incidence and sine of angle of refraction is constant. 

$\dfrac{\sin (i)}{\sin (r)}=n_{21}=\dfrac{n_{2}}{n_{1}}$

Here, i is the angle of incidence and r is the angle of refraction.

n21 is the refractive index of the medium 2 with respect to 1.

When light travels from rarer to denser, the light bends towards the normal. When it travels from denser to rarer, the light bends away from the normal. The amount of bending is governed by Snell’s law. This concept was necessary to understand the concept of apparent depth. Next, we will be discussing the relation between real depth and apparent depth.

Real Depth and Apparent Depth Derivation and Relation with Refractive Index

Real depth can be defined as the depth at which an object is placed in a beaker or liquid such that the observer and object are in the same medium. The propagation of light from the observer to the object or vice versa should be in the same medium. 

Apparent depth can be defined as the depth of the object measured by the observer when the object and observer are in different media. The propagation of light from the observer to the object or vice versa should be in a different medium. 

The real depth and apparent depth formula of refractive index can be mathematically derived. Consider a beaker filled with water and a square block placed at the bottom. Now, we measure the position of the square block in the empty glass, which will be the real depth given by the symbol ‘r’ in image 1. We will fill the beaker with a liquid and observe the apparent depth given by ha

The apparent depth will be given as

ha=hr-t.

For $\triangle A D C$ we see that

$\tan \left(\theta_{r}\right)=\dfrac{d}{h_{a}}$

For$\triangle A B C$ we see that

$\tan \left(\theta_{i}\right)=\dfrac{d}{h_{r}}$

From Snell’s law, we know

$n_{2} \sin \left(\theta_{i}\right)=n_{1} \sin \left(\theta_{r}\right)$

We will be assuming a paraxial limit; the human eye has a diameter of 4mm - 7mm and the ray can only make small angles with it. So, the Snell’s law equation can be written as 

follows:

$\begin{align} &n_{2} \tan \left(\theta_{i}\right)=n_{1} \tan \left(\theta_{r}\right) \\ &n_{2} \dfrac{d}{h_{r}}=n_{1} \dfrac{d}{h_{a}} \\ &\dfrac{n_{2}}{n_{1}}=\dfrac{h_{r}}{h_{a}} \end{align}$

Here, we see the real depth and apparent depth formula in terms of refractive index. Now, in the next section, we will be discussing other phenomena occurring due to apparent depth. 

Shift in Position and Lateral Magnification

The shift in position of the image with respect to the object can be easily  calculated. It is just the difference between the real depth minus the apparent depth. 

$t=h_{r}-h_{a}$

For derivation of lateral magnification:

Consider an object of height ‘t’ placed in the glass beaker as shown in Image 2. 

Distance of the image of N from interface is $=\dfrac{n_{2}}{n_{1}} h_{r}$

Distance of the image of N from interface is $=\dfrac{n_{2}}{n}\left(h_{r}-t\right)$

Then height of image observed will be $=\dfrac{n_{2}}{n_{1}} t$

Real Depth and Apparent Depth Numericals

Example 1: A vessel contains water upto the height of 30cm and above it, and oil up to another 20cm of height. The refractive indices of water and oil are 1.33 and 1.30, respectively. Find the apparent depth of the vessel.

Solution:

We know the formula relating real depth, apparent depth and refractive index.

So apparent depth for water and oil  is given by:

$\begin{align} &\dfrac{d_{w}}{d_{w^{\prime}}}=\dfrac{1.33}{1.30} \\ &d_{w^{\prime}}=\dfrac{d_{w} \times 1.3}{1.33}=\dfrac{30 \times 1.3}{1.33}=29.32 \end{align}$

$\begin{align} &\dfrac{d_{o}}{d_{o^{\prime}}}=\dfrac{1.33}{1.30} \\ &d_{o^{\prime}}=\dfrac{d_{o} \times 1.3}{1.33}=\dfrac{20 \times 1.3}{1.33}=19.55 \end{align}$

Total apparent depth is given by:

dw+d0=48.87

dw is the real depth in water.

dw’ is the apparent depth in water. 

do  is the real depth in oil.

do’ is the apparent depth in oil.

Example 2: A bucket of total height 50 cm is half filled with a liquid of refractive index 1.5 and half with another liquid of refractive index 2. Find the apparent depth of the bucket for an observer directly above the bucket.

Solution: 

Since there are two different media, the apparent depths due to both media have to be added. d1 is the depth of the liquid of refractive index 1.5 and d2 is the depth of liquid with refractive index 2. Since the liquids fill half of the bucket, both d1 and d2 will be 25cm. Now the apparent depth due to both media will be given as

$\begin{align} &d=d_{1}\left(\dfrac{n_{0}}{1.5}\right)+d_{2}\left(\dfrac{n_{0}}{2}\right) \\ &d=25\left(\dfrac{1}{1.5}\right)+25\left(\dfrac{1}{2}\right) \\ &d=29.16 \mathrm{~cm} \end{align}$

So, the apparent depth is 29.16cm.

Conclusion

Apparent depth can be said to be a lateral shift in the image of the object due to refraction. As discussed in the article, apparent and real depth are related through refractive index of the medium, where the object is present or what the observer is viewing. The propagation of light from different media leads to refraction and when an observer views the object, they observe a shift in the object from its original position. This phenomena can be seen in nature during sunrise and sunset. Even when the sun has not crossed the horizon, we can see the sun due to the effect of refraction. This can be described as an apparent shift in the position of its image, although the object is not there at position.