Question

The focal length $f$ of a mirror is given by $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$, where $u$ and $v$ represent object and image distances, respectively

• A. $\frac{\Delta f}{f}=\frac{\Delta u}{u}+\frac{\Delta v}{v}$
• B. $\frac{\Delta f}{f}=\frac{\Delta u}{v}+\frac{\Delta v}{u}$
• C. $\frac{\Delta f}{f}=\frac{\Delta u}{u}+\frac{\Delta v}{v}-\frac{\Delta (u+v)}{u+v}$
• D. $\frac{\Delta f}{f}=\frac{\Delta u}{u}+\frac{\Delta v}{v}+\frac{\Delta U}{u+v}+\frac{\Delta v}{u+v}$

Answer 1

The correct option is D $\frac{\Delta f}{f}=\frac{\Delta u}{u}+\frac{\Delta v}{v}+\frac{\Delta U}{u+v}+\frac{\Delta v}{u+v}$

The focal length $f$ of a mirror is given by

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

where $v=$ image distance

$u=$ object distance

from equation (1),

$f=\frac{uv}{u+v}=uv(u+v{)}^{-1}$

Taking log both sides we get

$logf=logu+logv+log(u+v{)}^{-1}$

$logf=logu+logv-log(u+v)$

Differentiating with respect to each variable

$\frac{\Delta f}{f}=\frac{\Delta u}{u}+\frac{\Delta v}{v}+\frac{\Delta (u+v)}{u+v}$

$\frac{\Delta f}{f}=\frac{\Delta u}{u}+\frac{\Delta v}{v}+\frac{\Delta u}{u+v}+\frac{\Delta v}{u+v}$

The correct option is D.