Question

For a concave mirror, if real image is formed the graph between $\frac{1}{u}$ and $\frac{1}{v}$ is of the form

• A.
• B.
• C.
• D.

Answer 1

Answer: (A)

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 $y=mx+c$
Slope $=m=\tan \theta =-1$
$\Rightarrow \theta =135^{\circ }$ or $-45^{\circ }$ and intercept $C=+\frac{1}{f}$