Question

At what distance from a convex lens should an object be placed to get an image of the same size as that of the object on a screen ?

• A. Beyond twice the focal length of the lens.
• B. At the principal focus of the lens.
• C. At twice the focal length of the lens.
• D. Between the optical centre of the lens and its principal focus.

Answer 1

The Correct Option is C

Step 1: Understanding the problem:
We are asked to find the distance at which an object should be placed in front of a convex lens to get an image of the same size as that of the object.
This indicates that the magnification of the lens must be 1 (i.e., the image and object sizes are equal).

Step 2: The magnification formula for a lens:
The magnification \( M \) of a lens is given by the formula:
\[ M = \frac{\text{Image Height}}{\text{Object Height}} = \frac{\text{Image Distance (v)}}{\text{Object Distance (u)}} \] Since the magnification is 1 (i.e., the image size equals the object size), we have:
\[ \frac{v}{u} = 1 \quad \Rightarrow \quad v = -u \] Thus, the image distance \( v \) is equal in magnitude but opposite in sign to the object distance \( u \).

Step 3: The lens formula:
The lens formula relates the focal length \( f \), object distance \( u \), and image distance \( v \) as:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] Substitute \( v = -u \) into the lens formula:
\[ \frac{1}{f} = \frac{1}{-u} + \frac{1}{u} \] \[ \frac{1}{f} = 0 \] This equation is not valid, which suggests that the relationship we used must involve a special case.

Step 4: The special case for magnification equal to 1:
When magnification is 1, the object and image distances are equal in magnitude and must be placed at a distance that is twice the focal length of the lens.
This is because at this position, the object and image are symmetrically placed on either side of the focal point.

Conclusion:
The object should be placed at a distance equal to twice the focal length of the convex lens, i.e., at \( 2f \), to get an image of the same size as the object.