The radius of curvature of a convex mirror used as a rear view mirror in a moving car is 2.0 m. A truck is coming from behind it at a distance of 3.5 m.
Calculate (a) position, and (b) size, of the image relative to the size of the truck. What will be the nature of the image?
R = 2m,
f = \(\frac{R}{2}\) = 1 m
u = -3.5 m
We know that
(a) \(\frac{1}{v}\) + \(\frac{1}{u}\) = \(\frac{1}{f}\)
⇒ \(\frac{1}{v}\) + \(\frac{1}{-3.5}\) = \(\frac{1}{1}\)
⇒ \(\frac{1}{v}\) = 1+ \(\frac{1}{3.5}\) = 1 + \(\frac{10}{35}\)
= 1 + \(\frac{2}{7}\) = \(\frac{9}{7}\)
∴ v = \(\frac{7}{9}\) = 0.77 m
So, the image is formed 0.77m behind the mirror.
(b) Now, m = -\(\frac{v}{u}\)
= -\(\frac{0.77}{-3.5}\) = \(\frac{\frac{7}{9}}{7}\)
= \(\frac{1}{4.5}\)
As m is positive, so image formed is virtual and erect.