A man sits in a parked car. The side-view mirror of the car is a convex mirror of radius of curvature 2 m. A jogger runs towards the car from behind. As the jogger moves from 19 m to 9 m from the mirror, the image of the jogger moves through
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 5 cm
For a convex mirror the focal length is positive and equal to \( R/2 \), so \( f=+1\ \mathrm{m} \). The jogger is in front of the mirror, so \( u \) is negative. The mirror equation gives \( v=fu/(u-f) \) for each position of the jogger.
\( \begin{aligned} v_1 &= \frac{(1)(-19)}{-19-1} \\ &= \frac{19}{20} \\ &= 0.95\ \mathrm{m} \end{aligned} \)
\( \begin{aligned} v_2 &= \frac{(1)(-9)}{-9-1} \\ &= \frac{9}{10} \\ &= 0.90\ \mathrm{m} \end{aligned} \)
\( \begin{aligned} \Delta v &= 0.95-0.90 \\ &= 0.05\ \mathrm{m} \\ &= 5\ \mathrm{cm} \end{aligned} \)
The jogger covers 10 m, but the image moves a very small distance towards the mirror. Hence, the image moves through 5 cm.
NCERT (2026-27):
Example 9.4 – “From the mirror equation, … we get”
\( \displaystyle v=\frac{fu}{u-f} \)
“For convex mirror, since R = 2 m, f = 1 m.”