Q. A object is placed at a distance of $f/2\ $from a convex lens of focal length $f$. The image will be
A) At one of the foci, virtual and double its size
B) Is greater than 1.5 but less than 2.0
C) At $2f,$virtual and erect
D) None of the above
Explanation
(a) $\frac{1}{v} - \frac{1}{- f/2} = \frac{1}{f}$
$\frac{1}{v} = \frac{1}{f} - \frac{2}{f} = - \frac{1}{f}$
or $v = - f$
Again, $m = \frac{v}{u} = \frac{- f}{- f/2} = 2$
Clearly, the image is virtual and double the size
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.