A 2 meter long scale with least count of 0.2 cm is used to measur… — Solution | JEE Physics

MCQMediumJEE 2023Significant Figures & Error Analysis

A $2 \, \text{meter}$ long scale with least count of $0.2 \, \text{cm}$ 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 $80 \, \text{cm}$ mark and $1 \, \text{m}$ mark, respectively. The image of the object pin on the other side of lens coincides with image pin that is kept at $180 \, \text{cm}$ mark. The % error in the estimation of focal length is:

A
$0.51$
B
$1.02$
C
$0.85$
D
$1.70$

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The object pin is at $80 \, \text{cm}$ , the convex lens is at $100 \, \text{cm}$ , and the image pin is at $180 \, \text{cm}$ . The least count of the scale is $0.2 \, \text{cm}$ .

Find: The percentage error in the focal length of the convex lens.

First determine the object distance and image distance:

U = 100 - 80 = 20 \, \text{cm}, \qquad V = 180 - 100 = 80 \, \text{cm}

Using the lens formula:

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

f = \frac{uv}{u+v}

Therefore,

f = \frac{20 \times 80}{20+80} = 16 \, \text{cm}

For error analysis, start from:

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

Differentiating:

\frac{Df}{f^2} = \frac{Du}{u^2} + \frac{Dv}{v^2}

Since each distance is obtained from the difference of two scale readings, the error in each is:

\Delta u = 0.4 \, \text{cm}, \qquad \Delta v = 0.4 \, \text{cm}

Now,

\frac{\Delta f}{f} = \left[ \frac{16 \times 0.4}{(80)^2} + \frac{16 \times 0.4}{(20)^2} \right]

So,

\frac{\Delta f}{f} = 16 \times 0.4 \left( \frac{17}{400} \right)

Hence,

\% \, \text{Error} = \frac{17 \times 0.4}{400} \times 100 = 1.7

Therefore, the percentage error in the estimation of focal length is $1.70$ , so the correct option is D.

Common mistakes

Using the least count $0.2 \, \text{cm}$ directly as the error in $u$ and $v$ is incorrect, because each distance is found by subtracting two scale readings. The errors add, so use $\Delta u = 0.4 \, \text{cm}$ and $\Delta v = 0.4 \, \text{cm}$ instead.
Taking the object distance as $80 \, \text{cm}$ and image distance as $180 \, \text{cm}$ is wrong. These are scale marks, not distances from the lens. First subtract the lens position to get $u = 20 \, \text{cm}$ and $v = 80 \, \text{cm}$ .
Applying percentage error directly to $f = \frac{uv}{u+v}$ without careful handling can lead to incorrect algebra. Use the relation $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ and then differentiate as shown in the solution.

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