Question

A small fish, $0.4\,{\text{m}}$ below the surface of the lake is viewed through a convex lens of focal length $3\,{\text{m}}$ . The lens is kept $0.2\,{\text{m}}$ above the water surface such that the fish uses the optical axis of the lens. Find the distance of the image of the fish as seen by the observer.

Answer 1

First of all, we will calculate the shift and find the apparent depth of the object by adding the shift to the actual depth. Then we will find the total object distance as the lens is placed above the water surface by $0.2\,{\text{m}}$ . Using lens formula, we will do the substitution of required values followed by manipulation to find the image distance.

Complete step by step answer:
In the given question, we are supplied the following data:
The fish is present at a depth of $0.4\,{\text{m}}$ from the surface of water.
The lens here used is a convex lens.
The focal length of the lens used is $3\,{\text{m}}$ .
Again, the lens is placed at a height of $0.2\,{\text{m}}$ above the water surface.
We are asked to find the distance of image formed, from the lens.

To begin with, we will first convert the units to C.G.S, as for the convenience. You can also switch to S.I units if you prefer.
We know,
$1\,{\text{m}} = 100\,{\text{cm}}$
So, we can write:

$$\Rightarrow 0.4\,{\text{m}} = 0.4 \times 100\,{\text{cm}} \\\ \Rightarrow 0.4\,{\text{m}} = 40\,{\text{cm}} \\\$$

Again,

$$\Rightarrow 0.2\,{\text{m}} = 0.2 \times 100\,{\text{cm}} \\\ \Rightarrow 0.2\,{\text{m}} = 20\,{\text{cm}} \\\$$

Further, we may write:

$$\Rightarrow 3\,{\text{m}} = 3 \times 100\,{\text{cm}} \\\ \Rightarrow 3\,{\text{m}} = 300\,{\text{cm}} \\\$$

First, we will calculate the shift of the image by the water, as the object is under water, it would not seem to be present at a depth of $40\,{\text{cm}}$ though.
If we can calculate the shift, we will get the apparent depth at which the object seems to be present.
Shift by water,
$s = d\left( {1 - \dfrac{1}{\mu }} \right)$ …… (1)
Where,
$s$ indicates the shift of the object due to refraction.
$d$ indicates the actual depth of the object.
$\mu$ indicates the refractive of water.

Now, we substitute the required values in the equation (1), we get:

$$\Rightarrow s = 40\left[ {1 - \dfrac{1}{{\left( {\dfrac{4}{3}} \right)}}} \right] \\\ \Rightarrow s = 40\left[ {1 - \dfrac{3}{4}} \right] \\\ \Rightarrow s = \dfrac{{40}}{4} \\\ \Rightarrow s = 10\,{\text{cm}} \\\$$

So, the apparent depth $\left( x \right)$ of the object becomes:

$$\Rightarrow x = \left( { - 40 + 10} \right)\,{\text{cm}} \\\ \Rightarrow x = - 30\,{\text{cm}} \\\$$

Since, the lens is placed $20\,{\text{cm}}$ above the water surface. So, the net object distance now becomes:

$$\Rightarrow u = - (20 + 30)\,{\text{cm}} \\\ \Rightarrow u = - 50\,{\text{cm}} \\\$$

We have from the lens formula,
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$ …… (2)

Substituting the required values in equation (2),

$$\Rightarrow \dfrac{1}{{300}} = \dfrac{1}{v} - \dfrac{1}{{( - 50)}} \\\ \Rightarrow \dfrac{1}{{300}} = \dfrac{1}{v} + \dfrac{1}{{50}} \\\ \Rightarrow \dfrac{1}{v} = \dfrac{1}{{300}} - \dfrac{1}{{50}} \\\ \Rightarrow \dfrac{1}{v} = \dfrac{{1 - 6}}{{300}} \\\ \Rightarrow \dfrac{1}{v} = \dfrac{{ - 5}}{{300}} \\\ \Rightarrow \dfrac{1}{v} = \dfrac{{ - 1}}{{60}} \\\ \Rightarrow v = - 60{\text{cm}} \\\$$

Hence, the image of the fish is formed on the same side of the fish, which is at a distance of $60{\text{cm}}$ from where the lens is. Image of the fish is superimposed on the real fish itself.

Note: In this problem, most of the students make mistakes while taking the object distance. Due to refraction, the depth of the object won’t be $40\,{\text{cm}}$ . Remember there will be a shift in the object distance. Again, the focal length of a convex lens is positive, as it is a converging lens, while negative in case of a concave lens as it is diverging in nature. Do take care of that.