Question

A lens of power \(5\,D\) forms a virtual image having magnification \(2.5\). Find the position of the object.

• A. \(12\,cm\)
• B. \(30\,cm\)
• C. \(8\,cm\)
• D. \(16\,cm\)

Hint:

For a virtual image formed by a convex lens, \[ m>1 \] and the object lies between optical centre and focus.

Answer 1

The Correct Option is A

Step 1: Find focal length. \[ P=\frac{1}{f} \] \[ 5=\frac{1}{f} \] \[ f=0.2\,m=20\,cm \]

Step 2: Use magnification relation. \[ m=\frac{v}{u} \] Given \[ m=2.5 \] Thus, \[ v=2.5u \]

Step 3: Apply lens formula. \[ \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \] Substituting, \[ \frac{1}{20} =\frac{1}{2.5u}-\frac{1}{u} \] \[ \frac{1}{20} =\frac{1-2.5}{2.5u} \] \[ \frac{1}{20} =\frac{-1.5}{2.5u} \] \[ u=-12\,cm \] Hence object distance is \[ \boxed{12\,cm} \] Therefore correct option is \[ \boxed{(A)} \]