Question

An object is placed at a distance of $\frac{f}{2}$ from a convex lens. The image will be formed:

• A. At $-f$, virtual and double its size.
• B. At $f$, real and inverted.
• C. At $-2f$, virtual and erect.
• D. At $2f$, real and erect.

Answer 1

The correct option is A

At $-f$, virtual and double its size.

Given $u=\frac{-f}{2}$
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$\Rightarrow \frac{1}{f}=\frac{1}{v}+\left(\frac{1}{\frac{f}{2}}\right)\Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{2}{f}$
$\Rightarrow \frac{1}{v}=\frac{-1}{f}$
So $v=-f$ i.e., the image will be formed at focus of the lens on the same side as that of the object.
Now magnification $m=\frac{v}{u}=\frac{-f}{\frac{-f}{2}}=+2$
So the image formed will be virtual with magnification +2.