Question

An optical bench, to measure the focal length of lens, is $1.5m$ long and on the bench marks are with spacing $\frac{1}{4}cm$. Now a lens is placed at $75cm$ and pin type object is placed at $45cm$ marks on the bench. If its image is formed at $135cm$ find maximum possible error in calculation of focal length.

Answer 1

Step 1. Given data with a diagram:

The distance of the pin from the lens is $$\begin{gathered} OP=u \\ =\left(75-45\right) \\ =30cm \end{gathered}$$

and $∆u=0.5$

The distance of the image from the lens is $$\begin{gathered} OB=v \\ =60cm \end{gathered}$$

and $∆v=0.5$

Step 2: Finding the maximum possible error:

Using the lens formula,

$\begin{array}{rcl}\frac{1}{v}-\frac{1}{u}& =& \frac{1}{f}............\left(1\right)\end{array}$

$\begin{array}{rcl}& \Rightarrow & \frac{1}{60}-\frac{1}{30}=\frac{1}{f}\\ & \Rightarrow & f=20cm\end{array}$

Differentiating $\left(1\right)$ on both the sides

We get,

$$\begin{gathered} -\frac{∆v}{v^2}+\frac{∆u}{u}=-\frac{∆f}{f^2} \\ \frac{∆f}{f^2}=\frac{∆v}{v}-\frac{∆u}{u^2} \\ \frac{∆f}{f}=20\left[\frac{1}{4{\left(60\right)}^2}-\frac{1}{4{\left(30\right)}^2}\right] \\ \frac{∆f}{f}=\frac{5}{720} \\ \frac{∆f}{f}\times 100=\frac{5}{720}\times 100 \\ =0.69\% \end{gathered}$$

Hence, the maximum possible error in the calculation of focal length is $0.69\%$.