An object is placed 50 cm in front of a concave mirror of focal length 10 cm. The object is then moved to 20 cm from the mirror. The ratio of the new image height to the earlier image height is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 4
The image height is \( h^{\prime}=mh \) for the same object. So the ratio of image heights equals the ratio of the magnifications. For each position, the mirror equation gives \( v \) and then \( m=-v/u \). Here \( f=-10\ \mathrm{cm} \).
For \( u_1=-50\ \mathrm{cm} \):
\( \begin{aligned} \frac{1}{v_1} &= -\frac{1}{10}+\frac{1}{50} \\ &= -\frac{4}{50} \end{aligned} \)
\( v_1=-12.5\ \mathrm{cm} \)
\( \begin{aligned} m_1 &= -\frac{v_1}{u_1} \\ &= -\frac{(-12.5)}{(-50)} \\ &= -\frac{1}{4} \end{aligned} \)
For \( u_2=-20\ \mathrm{cm} \):
\( \begin{aligned} \frac{1}{v_2} &= -\frac{1}{10}+\frac{1}{20} \\ &= -\frac{1}{20} \end{aligned} \)
\( v_2=-20\ \mathrm{cm} \)
\( \begin{aligned} m_2 &= -\frac{v_2}{u_2} \\ &= -\frac{(-20)}{(-20)} \\ &= -1 \end{aligned} \)
\( \begin{aligned} \frac{m_2}{m_1} &= \frac{-1}{-1/4} \\ &= 4 \end{aligned} \)
Both images are real and inverted. Hence, the new image is 4 times as tall as the earlier one.
NCERT (2026-27):
The mirror equation – “Dividing it by v, we get”
\( \displaystyle \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \)\( \qquad (9.7) \)
“This relation is known as the mirror equation.”
The mirror equation – “… so that”
\( \displaystyle m=\frac{h^{\prime}}{h}=-\frac{v}{u} \)\( \qquad (9.9) \)