Question

The focal length of a mirror is given by \[\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}\] where $u$ and $v$ represent object and image distance respectively, the maximum relative error in $f$ is?

Answer 1

Hint:In this above question we will calculate the relative error i.e., When measuring something, absolute error refers to the real amount by which you were off or erroneous. The absolute error is compared to the size of the thing you're measuring with relative error. To compute relative error, you must first determine the absolute error.

Complete step by step answer:
Let us solve the given equation for the maximum relative error,
\[\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v} \\
\Rightarrow \dfrac{1}{f} = \dfrac{{u + v}}{{uv}} \\ \]
An error in $f$ represents $\Delta f$ and we know that error is added in both numerator and denominator.
$\dfrac{{\Delta f}}{f} = \dfrac{{\Delta u}}{u} + \dfrac{{\Delta v}}{v} + \dfrac{{\Delta \left( {u + v} \right)}}{{u + v}} \\
\Rightarrow \dfrac{{\Delta f}}{f} = \dfrac{{\Delta u}}{u} + \dfrac{{\Delta v}}{v} + \dfrac{{\Delta u}}{{u + v}} + \dfrac{{\Delta v}}{{u + v}} \\ $
Now, the maximum relative error is,
$\Delta f = \bar f \pm \sqrt {{{\left( {\dfrac{{\Delta u}}{u}} \right)}^2} + {{\left( {\dfrac{{\Delta v}}{v}} \right)}^2} + {{\left( {\dfrac{{\Delta u}}{{u + v}}} \right)}^2} + {{\left( {\dfrac{{\Delta v}}{{u + v}}} \right)}^2}} $

Hence, the maximum relative error is $\bar f \pm \sqrt {{{\left( {\dfrac{{\Delta u}}{u}} \right)}^2} + {{\left( {\dfrac{{\Delta v}}{v}} \right)}^2} + {{\left( {\dfrac{{\Delta u}}{{u + v}}} \right)}^2} + {{\left( {\dfrac{{\Delta v}}{{u + v}}} \right)}^2}} $.

Note:Always provide context in the form of units. Make sure the audience understands the measurement units you're employing. The relative error, on the other hand, does not use units of measurement. A relative inaccuracy of 10%, for example, is stated as a fraction or a percentage.