A concave mirror forms a real image of an object kept at a distance of 9 cm from it. If the object is taken away further from the mirror by 6 cm, the image size is reduced to (1/4)th of its previous size. The focal length of the mirror is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → −7 cm
Use the Cartesian sign convention: distances measured against the incident light are negative, so the object distances and the focal length of the concave mirror are negative. For a mirror, the magnification in terms of f and u is
\( \displaystyle m = -\frac{v}{u} = \frac{f}{f - u} \)
\( u_1 = -9\,\mathrm{cm}, \; u_2 = -9 - 6 = -15\,\mathrm{cm} \)
\( \displaystyle m_1 = \frac{f}{f + 9}, \quad m_2 = \frac{f}{f + 15} \)
Both images are real, so \( m_1 \) and \( m_2 \) have the same sign, and \( m_2 = m_1/4 \):
\( \displaystyle \frac{f}{f + 15} = \frac{1}{4}\,\frac{f}{f + 9} \)
\( \displaystyle 4(f + 9) = f + 15 \)
\( \displaystyle 3f = -21 \Rightarrow f = -7\,\mathrm{cm} \)
Check: \( m_1 = -7/2 = -3.5 \) and \( m_2 = -7/8 \), so the image is indeed reduced to one-fourth.