Question

An erect image 3 times the size of the object is obtained with a concave mirror of radius of curvature 36 cm.What is the position of the object?

Answer 1

  magnification m  = - v/u = h'/h  = +3
       v = -3 u        actually  u is negative.  Hence v is positive.
       R = -36 cm    =>  f = R/2 = -18 cm , negative for a concave mirror.

       1/v + 1/u = 1/f,   mirror equation

        1/(-3u)  + 1/u = 1/(-18)
        2/(3u)  = -1/18
        
           u = - 12 cm    v = +36 cm 
  the object is placed 12 cm from pole of concave mirror ,  less than the focal length, to have an ERECT magnified image 3 times the object.
 

Answer 2

Answer:

hssh

$\cot(60)$