Question

An object $A B E D$ is placed in front of a concave mirror beyond centre of curvature $C$ as shown in figure. State the shape of the image.

• A. $\left|m_{A B}\right| < 1$ and $\left|m_{E D}\right| < 1$
• B. $\left|m_{A B}\right| > 1$ and $\left|m_{E D}\right| < 1$
• C. $\left|m_{A B}\right| < 1$ and $\left|m_{E D}\right| > 1$
• D. $\left|m_{A B}\right| > 1$ and $\left|m_{E D}\right| > 1$

Answer 1

Answer: (A)

Object is placed beyond $C .$ Hence, the image will be real and it will lie between $C$ and $F$. Further $u, v$ are negative, hence the mirror Formula will become
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}=\frac{u-f}{u f}$ or $v=\frac{f}{1-(f / u)}$

Now $u_{A B} > u_{E D}$
$\therefore v_{A B} < v_{E D}$
$\left|m_{A B}\right| < \left|m_{E D}\right|$
$\left(m=-\frac{v}{u}\right)$
Therefore, shape of the image will be as shown in figure. Also note that,
$v_{A B} < u_{A B}$ and $v_{E D} < u_{E D}$,
So, $\left|m_{A B}\right| < 1$ and $\left|m_{E D}\right| < 1$