Question

A short linear object of length b lies along the axis of a concave mirror of focal length $\mathrm{f}$ at a distance $\mathrm{u}$ from the pole of the mirror. The size of the image is approximately equal to:

• (A) $\mathrm{b}{\left(\frac{u-f}{f}\right)}^{\frac{1}{2}}$
• (B) $\mathrm{b}{\left(\frac{b}{u-f}\right)}^{\frac{1}{2}}$
• (C) $(\frac{u-f}{bf})$
• (D) $\frac{b{f}^2}{(u-f{)}^2}$

Answer 1

The Correct Option is D

Explanation:
The object is always placed in front of the mirror hence the object distance is taken as negative whereas the centre of curvature and focus lie in the front of the concave mirror so that the radius of curvature and the focal length is taken as negative in the case of the concave mirror.By using sign convention, from mirror equation, we get;$\frac{1}{v}-\frac{1}{u}=\frac{-1}{f}$or $v=\frac{fu}{f-u}$So $,\frac{\partial v}{\partial u}=\frac{\partial \left(\frac{fu}{f-u}\right)}{\partial x}$or $dv=\frac{f^2}{(u-f{)}^2}\times du$Now, $du=b($Given$)$So, $dv=\frac{b{f}^2}{(u-f{)}^2}=$ image size.Hence, the correct option is (D).