Question

an object $ABED$ is placed in front of a concave mirror beyond the center of curvature $C$ as shown in figure. State the shape of the image

• A. $|m_{AB}|<1$ and $|m_{ED}|<1$
• B. $|m_{AB}|>1$ and $|m_{ED}|<1$
• C. $|m_{AB}|<1$ and $|m_{ED}|>1$
• D. $|m_{AB}|>1$ and $|m_{ED}|>1$

Answer 1

The correct option is A $|m_{AB}|<1$ and $|m_{ED}|<1$

Object is placed beyond $C$. Hence, the image will be real and if will lie between $C$ and $F$. Further $u$, and $v$ are negative, hence the mirror formula will become
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}=\frac{u-f}{uf}$ or $v=\frac{f}{1-\left(f/u\right)}$
Now $u_{AB}>u_{ED}$
$\therefore v_{AB}>v_{ED}$
$|m_{AB}|<|m_{ED}|(asm=-\frac{v}{u})$
Therefore, shape of the image will be as shown in figure.
Also note that, $v_{AB}<u_{AB}$ and $v_{ED}<u_{ED}$,
So, $|m_{AB}|<1$ and $|m_{ED}|<1$.