A rod of length $10 \mathrm{~cm}$ lies along the principal axis of a concave mirror of focal length $10…

A rod of length $10 \mathrm{~cm}$ lies along the principal axis of a concave mirror of focal length $10 \mathrm{~cm}$ in such a way that its end closer to the pole is $20 \mathrm{~cm}$ away from the mirror. The length of the image is
  1. $10 \mathrm{~cm}$
  2. $15 \mathrm{~cm}$
  3. $2.5 \mathrm{~cm}$
  4. $5 \mathrm{~cm}$

Solution

By mirror formula, image distance of $A$ $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ $\begin{aligned} \frac{1}{v_A}+\frac{1}{u} & =\frac{1}{f} \\ \frac{1}{v_A}+\frac{1}{(-30)} & =\frac{1}{-10} \\ v_A & =-15 \mathrm{~cm} \end{aligned}$ Also image distance of $C$ $v_C=-20 \mathrm{~cm}$ The length of image $=\left|v_A-v_C\right|$ $=|-15-(-20)|$ $=5 \mathrm{~cm}$

Asked in: NEET 2012 (Mains)

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