Question

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

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

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
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}=\frac{u-f}{uf}$ or $v=\frac{f}{1-(f/u)}$

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