A single slit diffraction experiment is performed to determine the slit width using the equation, bd

Q: Why can't we just use the standard formula $\frac{\Delta b}{b} = \frac{\Delta D}{D} + \frac{\Delta d}{d}$?

The standard formula is derived using calculus approximations (differentials) which are only valid for very small changes, typically when percentage errors are less than $5\%$. Here, the error in $d$ is $20\%$, making the approximation highly inaccurate.

Q: How do we know the absolute errors for $D$ and $d$?

The problem states they are measured with scales of least count $1\text{ cm}$ and $1\text{ mm}$. In experimental physics, the maximum possible absolute error in a single reading is taken as the least count of the measuring instrument.

Q: Why are the errors for $\lambda$ and $m$ taken as zero?

The problem explicitly states that the values of $\lambda$ and $m$ are "known precisely". This means they are exact constants for this calculation, so their uncertainties are zero.

Q: Why do we calculate $b_{\text{max}}$ by taking $D_{\text{max}}$ and $d_{\text{min}}$?

To find the maximum possible error, we need the maximum possible value of $b$. Since $b$ is directly proportional to $D$ and inversely proportional to $d$, $b$ is maximized when the numerator $D$ is as large as possible and the denominator $d$ is as small as possible.

Q: The answer key gives two answers. Which one is better?

While the exam might have awarded marks for the approximate answer ($75.60$) due to common student habits, the exact calculation ($94.50$) is the scientifically rigorous and correct approach when dealing with large percentage errors. Always aim for the exact calculation if an error exceeds $5\%$.

Question

A single slit diffraction experiment is performed to determine the slit width using the equation, $\frac{bd}{D} = m\lambda$, where $b$ is the slit width, $D$ the shortest distance between the slit and the screen, $d$ the distance between the $m^{	ext{th}}$ diffraction maximum and the central maximum, and $\lambda$ is the wavelength. $D$ and $d$ are measured with scales of least count of $1	ext{ cm}$ and $1	ext{ mm}$, respectively. The values of $\lambda$ and $m$ are known precisely to be $600	ext{ nm}$ and $3$, respectively. The absolute error (in $\mu	ext{m}$) in the value of $b$ estimated using the diffraction maximum that occurs for $m = 3$ with $d = 5	ext{ mm}$ and $D = 1	ext{ m}$ is ___

Sigma Percentile · Accepted Answer

This problem is a beautiful example of how a standard, deeply ingrained habit can lead you straight into a trap. It tests not just your knowledge of optics, but your fundamental understanding of the mathematical tools used in error analysis.

### Analyzing the Setup

We are given a single slit diffraction experiment where the slit width $b$ is determined using the formula:

$$b = \frac{m\lambda D}{d}$$

Before we worry about errors, let's find the nominal value of $b$. We are given $D = 1	ext{ m}$, $d = 5	ext{ mm} = 5 	imes 10^{-3}	ext{ m}$, $\lambda = 600	ext{ nm} = 600 	imes 10^{-9}	ext{ m}$, and $m = 3$. Substituting these values:

$$b = \frac{3 	imes 600 	imes 10^{-9} 	imes 1}{5 	imes 10^{-3}} = 360 	imes 10^{-6}	ext{ m} = 360	ext{ }\mu	ext{m}$$

### The Standard (But Flawed) Approach

When asked to find the absolute error, the immediate instinct of almost every physics student is to use the fractional error formula derived from logarithmic differentiation:

$$\frac{\Delta b}{b} = \frac{\Delta m}{m} + \frac{\Delta \lambda}{\lambda} + \frac{\Delta D}{D} + \frac{\Delta d}{d}$$

Since $m$ and $\lambda$ are known precisely, their errors are zero. The absolute errors for $D$ and $d$ are their respective least counts: $\Delta D = 1	ext{ cm} = 0.01	ext{ m}$ and $\Delta d = 1	ext{ mm} = 10^{-3}	ext{ m}$. Plugging these in:

$$\frac{\Delta b}{b} = \frac{0.01}{1} + \frac{10^{-3}}{5 	imes 10^{-3}} = 0.01 + 0.20 = 0.21$$

This gives an absolute error of $\Delta b = 360 	imes 0.21 = 75.6	ext{ }\mu	ext{m}$.

### The Catch: When Approximations Fail

**Look closely at the fractional error of $d$:** $\frac{\Delta d}{d} = \frac{1}{5} = 0.20$, which is a massive **$20\%$ error!**

The standard fractional error formula is a first-order Taylor series approximation. It assumes that the changes (errors) are infinitesimally small. As a rule of thumb in physics, this approximation is only reliable when individual percentage errors are less than $5\%$. Because the error in $d$ is so large, the calculus-based approximation breaks down completely.

### The Rigorous Exact Calculation

To find the true maximum absolute error, we must abandon the approximation and calculate the exact maximum possible value of $b$. Looking at the formula $b = \frac{m\lambda D}{d}$, we can see that $b$ is maximized when the numerator is maximized and the denominator is minimized.

$$b_{	ext{max}} = \frac{m\lambda D_{	ext{max}}}{d_{	ext{min}}} = \frac{m\lambda (D + \Delta D)}{d - \Delta d}$$

Let's substitute the extreme values:

$$b_{	ext{max}} = \frac{3 	imes 600 	imes 10^{-9} 	imes (1 + 0.01)}{(5 - 1) 	imes 10^{-3}} = \frac{1800 	imes 10^{-9} 	imes 1.01}{4 	imes 10^{-3}}$$

$$b_{	ext{max}} = 450 	imes 1.01 	imes 10^{-6}	ext{ m} = 454.5	ext{ }\mu	ext{m}$$

### Final Conclusion

The exact maximum absolute error is simply the difference between this maximum possible value and our nominal value:

$$\Delta b_{	ext{max}} = b_{	ext{max}} - b = 454.5 - 360 = 94.5	ext{ }\mu	ext{m}$$

While an exam might occasionally accept the approximate answer, the exact calculation of $94.5	ext{ }\mu	ext{m}$ is the scientifically rigorous truth. Always check your percentage errors before blindly applying calculus approximations!

Enter Numerical Value:

Visualized Solution

\text{Visualizing the Setup}

\text{Extracting Given Values}

\text{Calculating Nominal Value of } b

\text{Standard Error Approximation}

\text{Approximate Absolute Error}

\text{The Trap: Large Percentage Error}

\text{Exact Maximum Value of } b

\text{Exact Absolute Error}

The Sigma Insight: Errors in Measurement

Analyzing the Setup

The Standard (But Flawed) Approach

The Catch

The Rigorous Exact Calculation

Final Conclusion

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