In an experiment for determination of the focal length of a thin convex lens, the distance of the ob

Q: Why do we use a negative sign for $u$ when the question says the distance is 10 cm?

In optics, we use the Cartesian sign convention. Distances measured in the direction of incident light are positive, and those opposite are negative. Since the object is placed in front of the lens, its coordinate is negative, hence $u = -10 \text{ cm}$.

Q: Why do we add the errors even though there is a minus sign in the differentiated equation?

Errors represent the maximum possible uncertainty. If we subtracted them, we might artificially calculate a smaller error than what is physically possible. To find the maximum permissible error, we always add the absolute values of the individual errors.

Q: Can I use the formula $\frac{\Delta f}{f} = \frac{\Delta u}{u} + \frac{\Delta v}{v}$ directly?

No! That formula only works for products and quotients (like $f = u \cdot v$). For equations involving addition or subtraction of reciprocals, you must differentiate the equation to find the correct error relationship.

Q: What happens if the image was virtual instead of real?

If the image were virtual, it would form on the same side as the object. By sign convention, $v$ would then be negative. You would substitute a negative value for $v$ in the lens formula, which would change the calculated focal length and subsequently the error.

Q: Why did we calculate $\Delta f$ first instead of directly finding the percentage error?

The differentiated equation naturally gives us an expression for $\frac{\Delta f}{f^2}$. While you can algebraically manipulate it to find $\frac{\Delta f}{f}$ directly, calculating $\Delta f$ first is often less prone to algebraic mistakes and gives you the absolute error as an intermediate check.

Question

In an experiment for determination of the focal length of a thin convex lens, the distance of the object from the lens is $10 \pm 0.1 \mathrm{\ cm}$ and the distance of its real image from the lens is $20 \pm 0.2 \mathrm{\ cm}$. The error in the determination of focal length of the lens is $n \%$. The value of $n$ is _______.

Sigma Percentile · Accepted Answer

The problem of finding the error in the focal length of a lens is a classic intersection of optics and error analysis. It tests not only your ability to apply the lens formula but also your understanding of how uncertainties propagate through non-linear equations. Let's break down the journey to the solution.

### Analyzing the Setup

Imagine you are in a physics lab, standing in front of an optical bench. You have a thin convex lens, an object pin, and an image screen. You measure the object distance to be $10 	ext{ cm}$ and the image distance to be $20 	ext{ cm}$. However, no measurement is perfect. Your ruler has a least count, leading to an uncertainty of $0.1 	ext{ cm}$ for the object distance and $0.2 	ext{ cm}$ for the image distance.

Before we even think about errors, we must establish our sign convention. In optics, distances measured in the direction of incident light are positive, while those measured against it are negative. Since the object is placed in front of the lens, its coordinate is negative:
$$u = -10 	ext{ cm}$$

The real image is formed on the other side of the lens, so its coordinate is positive:
$$v = +20 	ext{ cm}$$

### The Master Equation

To find the focal length $f$, we use the standard lens formula:
$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$

Substituting our values into this equation requires careful attention to the negative sign of $u$. A silly mistake here will derail the entire calculation.
$$\frac{1}{20} - \frac{1}{-10} = \frac{1}{f}$$

This simplifies beautifully:
$$\frac{1}{20} + \frac{1}{10} = \frac{3}{20}$$

Taking the reciprocal, we find the exact focal length:
$$f = \frac{20}{3} 	ext{ cm}$$

### The Calculus of Errors

Now comes the thrilling part: error propagation. We cannot simply add the percentage errors of $u$ and $v$ because they are not multiplied or divided; they are added as reciprocals. To find how the errors in $u$ and $v$ affect $f$, we must differentiate the lens formula.

Taking the differential of both sides:
$$-\frac{dv}{v^2} + \frac{du}{u^2} = -\frac{df}{f^2}$$

In error analysis, we are always interested in the **maximum permissible error**. This means we must assume the worst-case scenario where all errors compound. Therefore, we convert the differentials to absolute errors ($\Delta$) and force all terms to be positive:
$$\frac{\Delta f}{f^2} = \frac{\Delta v}{v^2} + \frac{\Delta u}{u^2}$$

Rearranging to isolate $\Delta f$, we get:
$$\Delta f = f^2 \left( \frac{\Delta v}{v^2} + \frac{\Delta u}{u^2} ight)$$

### Final Calculation

Now, we substitute our known values and their respective errors into this powerful equation. Notice that squaring the negative $u$ makes it positive, which aligns perfectly with our goal of finding the maximum error.
$$\Delta f = \left(\frac{20}{3}ight)^2 \left( \frac{0.2}{20^2} + \frac{0.1}{(-10)^2} ight)$$

Let's crunch the numbers:
$$\Delta f = \frac{400}{9} \left( \frac{0.2}{400} + \frac{0.1}{100} ight)$$

To add the fractions inside the bracket, we find a common denominator:
$$\Delta f = \frac{400}{9} \left( \frac{0.2 + 0.4}{400} ight)$$

The $400$ in the numerator and denominator cancel out elegantly, leaving us with:
$$\Delta f = \frac{0.6}{9} = \frac{1}{15} 	ext{ cm}$$

Finally, the question asks for the percentage error in the focal length, which is given by $\frac{\Delta f}{f} 	imes 100$.
$$	ext{\% Error} = \frac{1/15}{20/3} 	imes 100$$

This simplifies to:
$$	ext{\% Error} = \frac{1}{15} 	imes \frac{3}{20} 	imes 100 = \frac{3}{300} 	imes 100 = 1\%$$

Thus, the percentage error is exactly $1\%$, meaning our final answer for $n$ is **$1$**.

Enter Numerical Value:

Visualized Solution

Given Values of $u$ and $v$

The Lens Formula

Substituting Values

Calculating Focal Length $f$

Error Analysis via Differentiation

Maximum Absolute Error $Δ f$

Substituting Error Values

Calculating Absolute Error

Percentage Error

The Sigma Insight: Errors in Measurement

Analyzing the Setup

The Master Equation

The Calculus of Errors

Final Calculation

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