Where should an object be placed in front of a convex lens to get a real image of the size of the object?
A. At the principal focus of lens.
B. At twice of the focal length.
C. At infinity.
D. Between the optical centre of the lens and its principal focus.

Hint: As per the question we have to get the real image with the same size of object. Same size will be possible when magnification becomes one or negative of one. This is the key idea to solve this question.

Complete answer:
First of all we have to know the difference between virtual image and real image. A real image is defined as the collection of points of light rays which is coming from an object in which the rays are converging in nature. A virtual image is defined as the group of points of light rays which is coming from an object in which the rays are diverging in nature. Also a virtual will not make a visible projection on a screen.




Magnification is given by the equation,\[M=-1\]
$M=\dfrac{v}{u}=\dfrac{{{h}_{i}}}{{{h}_{o}}}$
Here

$\begin{align}
  & v=2F \\
 & u=-2F \\
\end{align}$

Therefore magnification becomes
\[M=-1\]

If an object is kept at double the focal length, then the image is being created at twice the focal length on the opposite side. So in this case magnification will be -1 and therefore the size of the image will be the same as the size of the object.

So, the correct answer is “Option B”.

Note:
Real images seem to be inverted always. By the way Virtual images are always seen as upright. Virtual image is the opposite of that of real image. We can call virtual images imaginary images. A reduced virtual image is formed by a single lens which is negative, independent of the position of the object.