Question

An object is placed at the distance of $\frac{f}{2}$ from convex lens, the image will be:
$f:\text{Focal Length}$

• A. At $f$, virtual and double the size.
• B. At $\frac{f}{2}$, real and double size
• C. At $2f$, virtual and double size
• D. At $f$, real and double size

Answer 1

The correct option is A At $f$, virtual and double the size.

Given:
Focal Length is $f$ (positive for convex lens).
Object distance, $u=-\frac{f}{2}$

Let image distance be $v$
Using lens formula,
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$\frac{1}{f}=\frac{1}{v}-\left(\frac{1}{-f/2}\right)$
$⟹v=-f$

Image is also formed on the left of the lens. Hence, it is virtual.

Magnification is given by:
$m=\frac{v}{u}=\frac{h^{\prime }}{h}$
$⟹\frac{-f}{-f/2}=\frac{h^{\prime }}{h}$
$h^{\prime }=2h$
Therefore, the image is double the size of the object.