A boy of height 1 m stands in front of a convex mirror. His distance from the mirror is equal to the focal length of the mirror, the height of the image is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 0.50\,\mathrm{m} \)
Use the mirror equation with the Cartesian sign convention: distances measured from the pole, positive in the direction of the reflected light (behind the mirror). The focal length of a convex mirror is positive, and the object in front of it has a negative distance.
\( f = +f \)
\( u = -f \)
\( h = 1\,\mathrm{m} \)
\( \displaystyle \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \)
\( \displaystyle \frac{1}{v} = \frac{1}{f} - \frac{1}{(-f)} = \frac{2}{f} \;\Rightarrow\; v = \frac{f}{2} \)
The magnification of a mirror is
\( \displaystyle m = -\frac{v}{u} = -\frac{f/2}{-f} = \frac{1}{2} \)
\( \displaystyle h' = m\,h = \frac{1}{2} \times 1\,\mathrm{m} = 0.50\,\mathrm{m} \)
The image is virtual and erect, half the height of the boy.