Q. An object $2.5\ cm$ high is placed at a distance of $10\ cm$ from a concave mirror of radius of curvature $30\ cm$. The size of the image is
A) 9.2 $cm$
B) 10.5 $cm$
C) 5.6 $cm$
D) 7.5 $cm$
Explanation
(d) $R = - 30cm\ \Rightarrow f = - 15cm$
$O = + 2.5cm,\ u = - 10\ cm$
By mirror formula $\frac{1}{- 15} = \frac{1}{v} + \frac{1}{( - 10)} \Rightarrow v = 30\ cm$
Also $\frac{I}{O} = - \frac{v}{u} \Rightarrow \frac{I}{( + 2.5)} = - \frac{30}{( - 10)} \Rightarrow I = + 7.5\ cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.