Question

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

• A. At one of the foci, virtual and double its size
• B. At $\frac{3f}{2}$, real and inverted
• C. At 2f, virtual and erect
• D. None of these

Answer 1

The correct option is A At one of the foci, virtual and double its size

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$ (Given $u=\frac{-f}{2}$ )
$\Rightarrow \frac{1}{f}=\frac{1}{v}+\left(\frac{1}{\frac{f}{2}}\right)\Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{2}{f}$
$\Rightarrow \frac{1}{v}=\frac{-1}{f}$ and $m=\frac{v}{u}=\frac{f}{\frac{f}{2}}$ = 2
So virtual at the focus and of double size.