Question

A small fish, $4\,{\text{cm}}$ below the surface of a lake, is viewed through t thin converging lens of focal length $30\,{\text{cm}}$ held $2\,{\text{cm}}$ above the water surface. Refractive index of water is $1.33$. The image of the fish from the lens is at a distance of:
A. $10\,{\text{cm}}$
B. $8\,{\text{cm}}$
C. $6\,{\text{cm}}$
D. $4\,{\text{cm}}$

Answer 1

Use the formula for apparent depth of an object. This formula gives the relation between refractive index of the medium in which object is placed, actual and apparent depth of the object. Also use lens formula. This formula gives the relation between the object distance from lens, image distance from lens and focal length of the lens. Hence, determine the apparent distance of fish and then determine the image distance of the fish.

Formulae used:
The refractive index $\mu$ of a medium is
$\mu = \dfrac{d}{{d'}}$ ……. (1)
Here, $d$ is the actual depth of an object and $d'$ is the apparent depth of an object.
The lens formula is given by
$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ …… (2)
Here, $v$ is the image distance, $u$ is the object distance and $f$ is the focal length of the lens.

Complete step by step answer:
We have given that the distance of the fish from the surface of water is $4\,{\text{cm}}$ and the distance of converging lens from the surface of water is $2\,{\text{cm}}$.
$d = 4\,{\text{cm}}$
$\Rightarrow{d_L} = 2\,{\text{cm}}$
The focal length of the converging lens is $30\,{\text{cm}}$.
$f = 30\,{\text{cm}}$
The refractive index of water is 1.33.
$\mu = 1.33$
We should first determine the apparent depth of the fish from the surface of water.
Rearrange equation (1) for apparent depth $d'$ of the fish from the surface of water.
$d' = \dfrac{d}{\mu }$

Substitute $4\,{\text{cm}}$ for $d$ and $1.33$ for $\mu$ in the above equation.
$d' = \dfrac{{4\,{\text{cm}}}}{{1.33}}$
$\Rightarrow d' = 3\,{\text{cm}}$
Hence, the apparent depth of the fish from the surface of water is $3\,{\text{cm}}$.
Thus, the distance of object from the converging lens is
$u = {d_L} + d'$
Substitute $2\,{\text{cm}}$ for ${d_L}$ and $3\,{\text{cm}}$ for $d'$ in the above equation.
$u = \left( {2\,{\text{cm}}} \right) + \left( {3\,{\text{cm}}} \right)$
$\Rightarrow u = 5\,{\text{cm}}$
Hence, the distance of the image from the converging lens is $5\,{\text{cm}}$.

Substitute $- 5\,{\text{cm}}$ for $u$ and $30\,{\text{cm}}$ for $f$ in equation (2).
$\dfrac{1}{v} - \dfrac{1}{{ - 5\,{\text{cm}}}} = \dfrac{1}{{30\,{\text{cm}}}}$
$\Rightarrow \dfrac{1}{v} = \dfrac{1}{{30}} - \dfrac{1}{5}$
$\Rightarrow \dfrac{1}{v} = \dfrac{{5 - 30}}{{150}}$
$\Rightarrow \dfrac{1}{v} = \dfrac{{ - 25}}{{150}}$
$\Rightarrow \dfrac{1}{v} = - \dfrac{1}{6}$
$\therefore v = - 6\,{\text{cm}}$
Therefore, the image of the fish will be formed at a distance $6\,{\text{cm}}$ from the converging lens.

Hence, the correct option is C.

Note: The students may think that focal length for the converging lens is positive and the object and image distance is negative. According to the sign conventions, the focal length of the converging lens is positive and the object and image distance on the same side of the converging lens is negative because distances on the front side of the converging lens are taken negative.