Q. The image of point $P$ when viewed from top of the slabs will be
A) 2.0 $cm$ above $P$
B) 1.5 $cm$ above $P$
C) 2.0 $cm$ below $P$
D) 1 $cm$ above $P$
Explanation
(d) The two slabs will shift the image a distance
$d = 2\left( 1 - \frac{1}{\mu} \right)t = 2\left( 1 - \frac{1}{1.5} \right)(1.5) = 1\ cm$
Therefore, final image will be $1\ cm$ above point $P$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.