Question

An object $\text{ABED}$ is placed in front of a concave mirror beyond the center of curvature $\text{C}$ as shown in figure. Which of the following options is true ?

• 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 $\text{C}$. Hence, the image will be real, and it lies between $\text{C}$ and $\text{F}$. Further $u$ and $v$ are negative for given condition.

Now, using mirror formula
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

we get,

Image distance $v=\frac{f}{1-\left(\frac{f}{u}\right)}.........\left(1\right)$

From Eq.$\left(1\right)$, we can say that, if $u\uparrow$ then $v\downarrow$


From the given diagram, it is clear that $u_{AB}>u_{ED}$ $\therefore v_{AB}<v_{ED}$

For magnification$\left(m\right)$

$|m_{AB}|<|m_{ED}|(\because m=-\frac{v}{u})$

Therefore, the shape of the image will be as shown in figure.


Also, from figure we can deduce that,

$v_{AB}<u_{AB}$ and $v_{ED}<u_{ED}$

So, $|m_{AB}|<1$ and $|m_{ED}|<1$.

Hence, option (a) is the correct alternative.