Discussion Forum

Consider a light source placed at a distance of 1.5 m along the axis facing the convex side of a spherical mirror of radius of curvature  1 m. The  position  (S'), nature and magnification  (m) of the image  are 

Answer:

Explanation:

 Given,

 Distance of light source (u) =-1.5 m=$-\frac{-3}{2}m$

 Radius of curvature  of mirror (R)= 1m

 we know that

   $\frac{1}{f}= \frac{1}{v}+\frac{1}{u}$          [ $\because$  $f=\frac{R}{2}$]

$\therefore$     $\frac{2}{R}=\frac{1}{v}+\frac{1}{u} \Rightarrow \frac{2}{1}= \frac{1}{v}+ \frac{1}{\frac{-3}{2}}$

   $\frac{2}{1}= \frac{1}{v}-\frac{2}{3} \Rightarrow  \frac{1}{v}=2+\frac{2}{3}$

 $\frac{1}{v}= \frac{6+2}{3} \Rightarrow \frac{1}{v}= \frac{8}{3} \Rightarrow v= \frac{3}{8}=0.375 m$

 The image is virtual

 Magnification (m)= $\frac{-v}{u}= \frac{\frac{-3}{8}}{\frac{-3}{2}}=\frac{1}{4}=0.25$