Question

An optical bench has $1.5\mathrm{m}$ long scale having four equal divisions in each $\mathrm{cm}$. While measuring the focal length of a convex lens, the lens is kept at $75\mathrm{cm}$ mark of the scale and the object pin is kept at $45\mathrm{cm}$ mark. The image of the object pin on the other side of the lens overlaps with image pin that is kept at $135\mathrm{cm}$ mark. In this experiment, the percentage error in the measurement of the focal length of the lens is ______.

Answer 1

Step 1. Given data

Optical bench $1.5\mathrm{m}$

The lens is kept at $75\mathrm{cm}$ mark, $x_2=75cm$

The object pin is kept at $45\mathrm{cm}$ mark, $x_1=45cm$

Step 2. Formula to be used.

We will use the lens formula is,

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

Here, $\mathrm{v}$ is the distance of the image from the lens, $\mathrm{u}$ is the distance of the object from the lens and $\mathrm{f}$ is the focal length of the lens.

Step 3. Calculate the percentage error in focal length.

From the above diagram,

The object distance is,

$$\begin{gathered} \left|\mathrm{u}\right|={\mathrm{x}}_2-{\mathrm{x}}_1 \\ \left|\mathrm{u}\right|=75-45 \\ \left|\mathrm{u}\right|=30\mathrm{cm} \end{gathered}$$

Since one centimeter is divided into four divisions, error in each measurement will be $\frac{1}{4}=0.25$

Error in $u$ = error in $75cm$ + error in $45cm$

$∆u=0.25+0.25=0.5cm$

The image distance is,

$$\begin{gathered} \left|\mathrm{v}\right|=135-75 \\ \left|\mathrm{v}\right|=60\mathrm{cm} \end{gathered}$$

Error in $v$ = error in $135cm$ + error in $75cm$

$∆v=0.25+0.25=0.5cm$

Now,

$$\begin{gathered} \frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \\ \frac{1}{60}-\frac{1}{-30}=\frac{1}{\mathrm{f}} \\ \frac{1}{60}+\frac{1}{30}=\frac{1}{\mathrm{f}} \\ f=20cm \end{gathered}$$ ("u" will be negative. By sign convention, distances on the left side are considered negative and the one on the right side of the lens is taken as positive)

The percentage error in focal length is,

$$\begin{gathered} \frac{\mathrm{dv}}{{\mathrm{v}}^2}+\frac{\mathrm{du}}{{\mathrm{u}}^2}=\frac{\mathrm{df}}{{\mathrm{f}}^2} \\ \frac{0.5}{{60}^2}+\frac{0.5}{{30}^2}=\frac{df}{20f} \\ \frac{1}{2\times 3600}+\frac{1}{2\times 900}=\frac{df}{20f} \\ \frac{df}{f}=\frac{20}{2}\left(\frac{1}{3600}+\frac{1}{900}\right) \\ \frac{df}{f}=\left(\frac{1}{360}+\frac{1}{90}\right) \\ \frac{df}{f}=0.0139 \\ \frac{df}{f}\times 100=0.0139\times 100 \\ \frac{df}{f}\times 100=1.39\% \end{gathered}$$

Percentage error $=1.39\%$

Thus, the percentage error in the measurement of the focal length of the lens is $1.39\%$.