Question

When an object is placed at a distance of $25cm$ from a spherical mirror, the magnification is $m_1$. When the object is moved $15cm$ away from mirror with respect to the earlier position, then magnification becomes $m_2$. If $\frac{m_1}{m_2}=4$, then the focal length of the mirror is :

• A. $-10cm$
• B. $20cm$
• C. $15cm$
• D. $-20cm$

Answer 1

The correct option is D $-20cm$

From the formula, $v=\frac{uf}{u-f}$

Magnification $m=\frac{-v}{u}=\frac{f}{f-u}$

From the question,

For first case, object distance $u_1=-25cm$

After the object is moved 15 cm away from the mirror, new object distance becomes $u_2=-40cm$

$\therefore$ $\frac{m_1}{m_2}=\frac{f-u_2}{f-u1}=4$

Or $\frac{f+40}{f+25}=4$

Or $3f=-60cm$

$⟹f=-20cm$