Question

A thin lens produces an upright image of the same size as the object. Find the distance of the object from the optical centre of the lens.
A) Zero
B) $4f$
C) $2f$
D) $\dfrac{f}{2}$

Answer 1

Hint: The size of the image is the same as that of the object. This implies that the magnification is one.

Formula used:
Thin lens formula is given by, $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ where $v$ is the image distance, $u$ is the object distance and $f$ is the focal length of the lens.
Magnification of the lens is given by, $m = \dfrac{v}{u}$ where $v$ is the image distance and $u$ is the object distance.

Complete step by step answer:
Step 1: Using the thin lens formula obtain an expression for the distance at which the image is formed.
Thin lens formula is given by, $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ -------- (1)
where $v$ is the image distance, $u$ is the object distance and $f$ is the focal length of the lens.
Equation (1) can be rearranged as $\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u}$ or, $\dfrac{1}{v} = \dfrac{{u - f}}{{fu}}$
By taking the reciprocal of the above equation we get, $v = \dfrac{{uf}}{{u - f}}$
Thus the distance at which the image is formed is obtained as $v = \dfrac{{uf}}{{u - f}}$ ------ (2).

Step 2: Express the relation for the magnification produced by the lens.
Magnification produced by a lens refers to the extent to which an object is magnified. It is the ratio of the distance at which the image is formed (image distance) to the distance at which the object is placed.
Magnification is given by, $m = \dfrac{v}{u}$ -------- (3) ,
where $v = \dfrac{{uf}}{{u - f}}$ is the image distance and $u$ is the object distance.
Substituting equation (2) in equation (3) we get, $m = \dfrac{{uf}}{{u\left( {u - f} \right)}}$
By cancelling out the similar terms in the above equation, we get $m = \dfrac{f}{{\left( {u - f} \right)}}$
Thus the magnification produced by the lens is $m = \dfrac{f}{{\left( {u - f} \right)}}$ ------- (4).

Step 3: Using equation (4) find the distance at which the object is placed.
It is given that the image formed is upright and has the same size as that of the object. So, magnification $m = 1$ .
Substituting the value of $m = 1$ in equation (4) we get, $\dfrac{f}{{\left( {u - f} \right)}} = 1$
Reduce the above equation to get $u - f = f$ or, $u = f - f = 0$

Therefore the object distance is $u = 0$

Note:
The distance at which the object is placed from the optical centre is calculated as zero. This essentially means that the object must be placed at the optical centre itself. This position of the object will then result in the formation of an image that has the same size as that of the object.