Question

An object is placed at a distance of $100cm$ from a concave lens of focal length $40cm$. Find the distance of the position of image from the lens in $cm$.

• A. $0.035$
• B. $28.57$
• C. $0.015$
• D. None

Answer 1

The correct option is B $28.57$

By using sign convention rules
Object distance, $u=-100cm$

Focal length of concave lens, $f=-40cm$

Using lens formula,

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
We have,

$\Rightarrow \frac{1}{-40}=\frac{1}{v}-\frac{1}{-100}$

$\Rightarrow \frac{1}{v}=\frac{1}{-40}-\frac{1}{100}$

$\Rightarrow \frac{1}{v}=\frac{-100-40}{4000}=\frac{-140}{4000}=\frac{-7}{200}$

$\Rightarrow v=\frac{-200}{7}=-28.57cm$

Magnification .

$m=\frac{v}{u}$ = $\frac{-28.57}{-100}=0.2857$

so magnification positive means virtual image formed.
"- " Indicates virtual image.
A virtual image is formed at a distance of $28.57cm$ from the optical center on the same side of the object.

Hence, the correct answer is OPTION B.