Thank you for reporting, we will resolve it shortly
Q.
For a concave mirror, if real image is formed the graph between $\frac{1}{u}$ and $\frac{1}{v}$ is of the form
Ray Optics and Optical Instruments
Solution:
Since $\frac{1}{f}=+\frac{1}{v}+\frac{1}{u} \Rightarrow \frac{1}{v}=-\frac{1}{u}+\frac{1}{f}$
Putting the sign convention properly
$\frac{1}{(-v)}=\frac{-1}{(-u)}+\frac{1}{(-f)} $
$\Rightarrow \frac{1}{v}=-\frac{1}{u}+\frac{1}{f}$
Comparing this equation with $y=m x+c$
Slope $=m=\tan \theta=-1 $
$\Rightarrow \theta=135^{\circ}$ or $-45^{\circ}$ and intercept $C=+\frac{1}{f}$
