Given:
Magnification (m) = 3 (virtual image is erect).
Radius of the curvature, R =36 cm
Focal Length (f) = \(\frac{R}2\) = -18cm (Negative for concave mirror).
Now by formula,
1. Magnification, m = \(\frac{-v}u\)
Where,
v is the image distance,
u is the object distance,
Putting the values in the above formula, we get
3 = \(\frac{-v}u\) or v = - 3u
2. Mirror formula
\(\frac{1}{f}=\frac1v+\frac1u\)
Where,
f is the focal length of the mirror
v is the image distance from the mirror
u is the object distance from the mirror
Thus, putting the values in the equation, we get
\(\frac{-1}{18}=\frac{-1}{3u}+\frac1u=\frac{2}{3u}\)
Thus, on solving the above equation, we get \(\frac{-1}{36}=\frac{1}{3u}\)
u=-12 cm
Hence the distance of the object from the mirror is 12 cm.
