An optical bench has 1.5 m long scale having four equal divisions in each cm. While measuring the fo

Q: Why is the error in $u$ and $v$ taken as $0.5 \text{ cm}$ and not $0.25 \text{ cm}$?

The object distance $u$ is found by subtracting the position of the object pin from the position of the lens ($u = x_{\text{lens}} - x_{\text{object}}$). Since both positions are measured using the scale, each reading has an error of $\pm 0.25 \text{ cm}$. When subtracting two quantities, their absolute errors add up, giving $\Delta u = 0.25 + 0.25 = 0.5 \text{ cm}$.

Q: Why do we add the error terms after differentiating the lens formula?

In error analysis, we always calculate the maximum possible error. Even though the derivative gives a negative sign for some terms, errors can occur in either direction. To ensure we account for the worst-case scenario, we take the absolute values of all terms and add them together.

Q: Do we need to use sign convention when substituting $u$ and $v$ into the error formula?

No, when calculating errors using the differentiated formula $\frac{\Delta f}{f^2} = \frac{\Delta v}{v^2} + \frac{\Delta u}{u^2}$, we only use the magnitudes of $u$ and $v$. The squares in the denominator naturally remove any negative signs, but conceptually, error is always a positive magnitude.

Q: What if the lens was fixed at the zero mark of the scale?

If the lens was permanently fixed exactly at the zero mark, its position would have no measurement error ($\Delta x_{\text{lens}} = 0$). In that case, the error in $u$ would just be the error in reading the object pin's position, so $\Delta u$ would be $0.25 \text{ cm}$.

Q: Can we calculate the error by finding $f_{\text{max}}$ and $f_{\text{min}}$ directly?

Yes, you could calculate $f_{\text{max}}$ using $u_{\text{min}}$ and $v_{\text{max}}$, and $f_{\text{min}}$ using $u_{\text{max}}$ and $v_{\text{min}}$, and then find the percentage difference. However, the calculus method (differentiation) is much faster, less prone to calculation mistakes, and is the standard expected approach for small errors in JEE.

Question

An optical bench has 1.5 m long scale having four equal divisions in each cm. While measuring the focal length of a convex lens, the lens is kept at 75 cm mark of the scale and the object pin is kept at 45 cm mark. The image of the object pin on the other side of the lens overlaps with image pin that is kept at 135 cm mark. In this experiment, the percentage error in the measurement of the focal length of the lens is________. [JEE(Advanced) 2019]

Sigma Percentile · Accepted Answer

### The Experimental Setup
Imagine you are in a physics lab, standing in front of a long optical bench. The bench has a scale, and mounted on it are three key components: an object pin, a convex lens, and an image pin. The goal of this classic experiment is to determine the focal length of the convex lens.

The problem states that the lens is placed at the $75 	ext{ cm}$ mark, the object pin is at the $45 	ext{ cm}$ mark, and the image pin is at the $135 	ext{ cm}$ mark. The fact that the image of the object pin perfectly overlaps with the image pin means there is no parallax—we have found the exact position of the real image!

### Decoding the Least Count
Before we jump into the optics, we need to understand the precision of our measuring instrument. The scale on the optical bench has four equal divisions in each centimeter.

What does this mean for our measurements? The smallest distance we can measure, known as the **Least Count (LC)**, is simply $1 	ext{ cm}$ divided by $4$.

$$ 	ext{LC} = \frac{1}{4} 	ext{ cm} = 0.25 	ext{ cm} $$

This $0.25 	ext{ cm}$ is the maximum possible error for any single reading taken on this scale.

### Calculating Distances and Their Errors
In optics, we don't just take single readings; we measure distances between two points. 
The object distance $u$ is the distance between the lens and the object pin:

$$ u = 75 	ext{ cm} - 45 	ext{ cm} = 30 	ext{ cm} $$

But here is where many students make a silly mistake! Since $u$ is calculated by subtracting two independent scale readings (the lens position and the object pin position), the errors in both readings will add up.

$$ \Delta u = \Delta x_{	ext{lens}} + \Delta x_{	ext{object}} = 0.25 	ext{ cm} + 0.25 	ext{ cm} = 0.5 	ext{ cm} $$

Similarly, the image distance $v$ is the distance between the image pin and the lens:

$$ v = 135 	ext{ cm} - 75 	ext{ cm} = 60 	ext{ cm} $$

And just like before, the error in $v$ is the sum of the errors of the two readings:

$$ \Delta v = \Delta x_{	ext{image}} + \Delta x_{	ext{lens}} = 0.25 	ext{ cm} + 0.25 	ext{ cm} = 0.5 	ext{ cm} $$

### The Master Equation: Lens Formula
Now that we have our distances, let's find the focal length $f$. We use the standard thin lens formula:

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

Applying the Cartesian sign convention, the object distance $u$ is negative ($-30 	ext{ cm}$) because it is measured against the direction of incident light, while the image distance $v$ is positive ($+60 	ext{ cm}$).

$$ \frac{1}{f} = \frac{1}{60} - \frac{1}{-30} = \frac{1}{60} + \frac{1}{30} $$

$$ \frac{1}{f} = \frac{1 + 2}{60} = \frac{3}{60} = \frac{1}{20} $$

So, the focal length $f$ is exactly $20 	ext{ cm}$.

### Error Analysis: The Calculus Connection
Finding the focal length was the easy part. Now comes the true JEE Advanced challenge: finding the percentage error in $f$. To do this, we differentiate the lens formula.

$$ d\left(\frac{1}{f}ight) = d\left(\frac{1}{v}ight) - d\left(\frac{1}{u}ight) $$

$$ -\frac{df}{f^2} = -\frac{dv}{v^2} + \frac{du}{u^2} $$

In error analysis, we always want to find the **maximum possible error**. Therefore, we take the absolute values and add the fractional error terms together:

$$ \frac{\Delta f}{f^2} = \frac{\Delta v}{v^2} + \frac{\Delta u}{u^2} $$

Rearranging this to find the fractional error in $f$:

$$ \frac{\Delta f}{f} = f \left( \frac{\Delta v}{v^2} + \frac{\Delta u}{u^2} ight) $$

### The Final Calculation
We are in the endgame now. Let's substitute all our known values into the error equation and multiply by $100$ to get the percentage error.

$$ 	ext{Percentage Error} = \frac{\Delta f}{f} 	imes 100 = f \left( \frac{\Delta v}{v^2} + \frac{\Delta u}{u^2} ight) 	imes 100 $$

$$ 	ext{Percentage Error} = 20 \left( \frac{0.5}{60^2} + \frac{0.5}{30^2} ight) 	imes 100 $$

$$ 	ext{Percentage Error} = 2000 \left( \frac{0.5}{3600} + \frac{0.5}{900} ight) $$

To add these fractions easily, let's make the denominators the same:

$$ 	ext{Percentage Error} = 2000 \left( \frac{0.5}{3600} + \frac{2.0}{3600} ight) = 2000 \left( \frac{2.5}{3600} ight) $$

$$ 	ext{Percentage Error} = \frac{5000}{3600} = \frac{50}{36} \approx 1.388... \% $$

Rounding off to two decimal places, we get **$1.39 \%$**.

This problem is a beautiful reminder of why we must be careful with measurements. A tiny $0.25 	ext{ cm}$ uncertainty on a scale propagates through our equations to create a $1.39 \%$ error in our final result!

Enter Numerical Value:

Visualized Solution

\text{Experimental Setup}

\text{Least Count of the Scale}

\text{Object Distance } (u) \text{ and its Error } (\Delta u)

\text{Image Distance } (v) \text{ and its Error } (\Delta v)

\text{Focal Length } (f)

\text{Error in Focal Length}

\text{Percentage Error Calculation}

\text{Final Answer}

The Sigma Insight: Errors in Measurement

The Experimental Setup Imagine you are in a physics lab, standing in front of a long optical bench

Decoding the Least Count Before we jump into the optics, we need to understand the precision of our measuring instrument

Calculating Distances and Their Errors In optics, we don't just take single readings; we measure distances between two points

The Master Equation

Error Analysis

The Final Calculation We are in the endgame now

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