Physics - Geometrical Optics Question with Solution | TestHub

The power of a lens is given by $P=\frac{1}{f}$

Using the lens formula, we can write the power in terms of object distance $u$ and image distance $v$ as:

$P=\frac{1}{v}-\frac{1}{u}$

Differentiating both sides to find the error in power, we get:

$dP=-\frac{dv}{v^2}+\frac{du}{u^2}$

To find the maximum permissible error, we take the absolute values of the individual error terms and add them:

$\Delta P=\frac{\Delta v}{v^2}+\frac{\Delta u}{u^2}$

Answer: $\frac{\Delta v}{v^2}+\frac{\Delta u}{u^2}$