Question

Consider a light source placed at a distance of $1.5m$ along the axis facing the convex side of a spherical mirror of radius of curvature $1m$. The position ($s^{\prime }$), nature and magnification ($m$) of the image are

• A. $s^{\prime }=0.375m$, Virtual, upright, $m=0.25$
• B. $s^{\prime }=0.375m$, Real, inverted, $m=0.25$
• C. $s^{\prime }=3.75m$, Virtual, upright, $m=2.5$
• D. $s^{\prime }=3.75m$, Real, upright, $m=2.5$

Answer 1

The correct option is A $s^{\prime }=0.375m$, Virtual, upright, $m=0.25$

Using mirror formula :
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}=\frac{2}{R}$
Given : $u=-1.5m$ $R=1m$
$\therefore \frac{1}{v}+\frac{1}{-1.5}=\frac{2}{1}$
$\Rightarrow v=0.375m=3/8m$
Since, $v>0$. Thus virtual image is formed at a distance $0.375m$ behind the mirror.
Magnification $m=\frac{-v}{u}=\frac{-3/8}{-3/2}=\frac{1}{4}=0.25$
Since, $m$ is positive, so an upright image is formed.