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  • A 2 meter long scale with least count of 0.2 cm is used to measure the locations of objects on an optical bench. While measuring the focal length of a convex lens, the object pin and the con

A 2 meter long scale with least count of 0.2 cm is used to measure the locations of objects on an optical bench.
While measuring the focal length of a convex lens, the object pin and the convex lens are placed at 80 cm mark
and 1 m mark., respectively. The image of the object pin on the other side of lens coincides with image pin that
is kept at 180 cm mark. The % error in the estimation of focal length is :

Option: 1

0.51 


Option: 2

 1.02


Option: 3

0.85


Option: 4

1.70


Answers (1)

Based on the data provided
U = 100 – 80 = 20 cm
V = 180 – 100 = 80 cm

Using   \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}} \quad \text { or } \mathrm{f}=\frac{\mathrm{uv}}{\mathrm{u}+\mathrm{v}}=\frac{20 \times 80}{20+80} \text { or } \mathrm{f}=16 \mathrm{~cm}

For error analysis,

\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}} \quad

Differentiating

-\frac{D f}{f^2}=-\frac{D v}{v^2}-\frac{\Delta u}{u^2}

To calculat e \Deltau &  \Deltav
U = (100 ± 2) – (80 ± 0.2) = (20 ± 0.4) c

Therefore \Deltau = 0.4 cm,                                                    
Similarly  \Deltav = 0.4 cm.

\begin{aligned} & \text { Now } \frac{\Delta \mathrm{f}}{\mathrm{f}}=\mathrm{f}\left[\frac{\Delta \mathrm{v}}{\mathrm{v}^2}+\frac{\Delta \mathrm{u}}{\mathrm{u}^2}\right] \\ & \frac{\Delta \mathrm{f}}{\mathrm{f}}=16\left[\frac{0.4}{(80)^2}+\frac{0.4}{(20)^2}\right] \end{aligned}  

 

(Note: every data is in cm)

\begin{aligned} & \frac{\Delta \mathrm{f}}{\mathrm{f}}=\frac{16 \times 0.4}{(20)^2}\left[\frac{1}{4^2}+1\right] \\ & =\frac{16 \times 0.4}{20^2} \times \frac{17}{16}=\frac{17 \times 0.4}{400} \\ & \% \text { Error : } \frac{\Delta \mathrm{f}}{\mathrm{f}} \times 100=\frac{17 \times 0.4}{400} \times 1000 \\ & =1.7 \end{aligned}

Posted by

Sumit Saini

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