Question

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

Answer 1

The image so produced by the thin lens is of the same size as the object. This gives us a hint that the magnification is one. If the magnification is one then it will be very easy to calculate the image distance from the lens.

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

Complete step by step answer:
It is given in the problem that a thin lens produces an image of the same size as the object and we need to find the object distance.
The lens formula is given by,
$\Rightarrow \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
On solving the expression of lens formula, we get
$\Rightarrow \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
$\Rightarrow \dfrac{1}{f} + \dfrac{1}{u} = \dfrac{1}{v}$
$\Rightarrow \dfrac{{u + f}}{{fu}} = \dfrac{1}{v}$
Taking the reciprocal of the equation, we get
$\Rightarrow v = \dfrac{{uf}}{{u + f}}$.........eq. (1)
From this we get the image distance.
Now, we will calculate the magnification.
$\Rightarrow m = \dfrac{v}{u}$
Where magnification is m the image distance is v and the object distance is u.
Putting the value of equation (1) in the magnification formula, we get.
$\Rightarrow m = - \dfrac{v}{u}$
$\Rightarrow m = \dfrac{{\dfrac{{uf}}{{u + f}}}}{u}$
$\Rightarrow m = \dfrac{{uf}}{{u(u + f)}}$
$\Rightarrow m = \dfrac{f}{{u + f}}$………eq. (2)
Thus, magnification produced by the lens is given by the equation (2).
We will now find the object distance form equation (2).
As we know, the image size and object size are the same, therefore, the magnification will be one.
If we substitute the value of magnification in the equation (2), we will get the object distance.
$\Rightarrow m = \dfrac{f}{{u + f}}$
$\Rightarrow 1 = \dfrac{f}{{u + f}}$
$\Rightarrow u + f = f$
$\Rightarrow u = f - f$
$\Rightarrow u = 0$

Therefore, Option A is the correct answer. Thus, the object distance (u) from the lens is zero. This means the object is placed at the optical centre of the lens.

Note:
-If the placement of the object at the optical centre of the lens then it forms the image of the same size as that of the object.
-Magnification refers to the extent up to which an object is magnified when placed before a lens.