Question

An object is placed 30 cm in front of a concave mirror of radius of curvature 40 cm. Find the (i) position of the image formed and (ii) magnification of the image.

Hint:

For mirrors, use the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ and apply the sign convention carefully for object and image distances.

Answer 1

(i) The lens formula is given by: \[ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \] Substituting the given values: \[ \frac{1}{v} + \frac{1}{-30} = \frac{1}{-20} \] Now, solving for \(v\): \[ \frac{1}{v} = \frac{1}{-20} - \frac{1}{-30} \] \[ \frac{1}{v} = \frac{-3 + 2}{60} = \frac{-1}{60} \] \[ v = -60 \, \text{cm} \] Thus, the value of \(v\) is \(-60 \, \text{cm}\). (ii) The magnification (\(m\)) is given by: \[ m = - \frac{v}{u} \] Substituting the values of \(v = -60 \, \text{cm}\) and \(u = -30 \, \text{cm}\): \[ m = - \left( \frac{-60}{-30} \right) \] \[ m = -2 \] Thus, the magnification is \(m = -2\).