Question

Which of the following graphs correctly represents the variation of $\frac{1}{v}$ versus $\frac{1}{u}$ for a concave mirror?
[$v$ = image distance, $u$ = object distance]

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

From the mirror formula,

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}.....\left(1\right)$

For a concave mirror, object distance $u$, image distance $v$ and focal length $f$ are all negative when object is between $\infty$ and focus.

$\therefore$ From $\left(1\right)$, we can write that

$\frac{1}{v}=-\frac{1}{u}+\frac{1}{f}$

Comparing this with $y=mx+c$

we get, $m=-1$ and $c=\frac{1}{f}$.
It represents a straight line graph with negative slope and negative intercept as shown below.

Also by convention, we take $u$ as negative and $v$ will be negative for real image (object between $\infty$ and focus) and positive for virtual image (object between focus and pole). Hence, the graph will be as follows.


Hence, option (c) is correct.