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Q.
A real inverted image in a concave mirror is represented by graph ( $\text{u} , \text{v} , \text{f}$ are coordinates)
NTA AbhyasNTA Abhyas 2020Ray Optics and Optical Instruments
Solution:
For real inverted image formed by concave mirror.
$\text{v} = - \text{v} \text{e} , \, \, \, \text{u} = - \text{v} \text{e} , \, \, \text{f} = - \text{v} \text{e}$
$⇒ \, \, \, \frac{u}{f} $ & $\frac{v}{f}$ are positive
So graph show be in 1st qudarant
$\text{and} \frac{1}{\text{v}} + \frac{1}{\text{u}} = \frac{1}{\text{f}}$
$⇒ \, \, \, \frac{\text{f}}{\text{v}} + \frac{\text{f}}{\text{u}} = 1$
This graph can also be realized by Newtons formula $\text{x}_{1} \text{x}_{2} = \text{f}^{2}$ where $\text{x}_{1} \text{,} \text{ and } \text{x}_{2}$ are distance from focus so graph will be rectangular hyperbola
$\Rightarrow $ (1) is right answer.