In a single slit diffraction experiment, a slit of width (0.016 ± 0.002) mm is used to measure the w

Q: Why do we use logarithmic differentiation here?

Taking the natural logarithm of an equation that involves products or quotients turns them into sums and differences. When we differentiate this new equation, it directly gives us the fractional errors (like $\frac{d\lambda}{\lambda}$), which is exactly what we need!

Q: Why must $\Delta\theta$ be in radians?

In calculus, the derivative of $\sin\theta$ is $\cos\theta$ *only* if $\theta$ is measured in radians. If you use degrees, an extra scaling factor of $\frac{\pi}{180}$ appears. To keep the standard formulas valid, any standalone $d\theta$ or $\Delta\theta$ must be in radians.

Q: How did we get $\cot\theta$ in the error equation?

When we differentiate $\ln(\sin\theta)$, we use the chain rule. The derivative of $\ln(x)$ is $\frac{1}{x}$, so we get $\frac{1}{\sin\theta}$. Then we multiply by the derivative of the inside function, $\sin\theta$, which is $\cos\theta$. This gives $\frac{\cos\theta}{\sin\theta}$, which is $\cot\theta$.

Q: Why do we add the errors instead of subtracting them?

When calculating maximum possible error, we always add the absolute values of the individual fractional errors. Errors can compound in the worst-case scenario, so we assume they add up to give the maximum uncertainty.

Q: Is it okay to approximate $\sqrt{1 - \sin^2\theta}$ as 1?

Yes! Since $\sin(2^\circ) = 0.035$ is very small, its square is $0.001225$. Subtracting this from 1 gives $0.998775$, and its square root is extremely close to 1. In competitive exams, making this approximation saves time without significantly affecting the final answer.

Question

In a single slit diffraction experiment, a slit of width $(0.016 \pm 0.002) 	ext{ mm}$ is used to measure the wavelength of a monochromatic light source. In the diffraction pattern, the angular distance between the central maximum and first minimum is measured to be $(2^\circ \pm 40')$. The value of the fractional error in the measurement of wavelength is: [Given: $\sin(2^\circ)= 0.035$]

Sigma Percentile · Accepted Answer

The problem of finding the fractional error in a single slit diffraction experiment is a classic test of both your understanding of wave optics and your mastery of error analysis. It’s not just about plugging numbers into a formula; it’s about carefully navigating the mathematical traps that examiners love to set.

### The Setup: Single Slit Diffraction

Imagine a beam of monochromatic light passing through a narrow slit of width $d$. As the light waves squeeze through the opening, they spread out and interfere with each other, creating a beautiful pattern of bright and dark fringes on a distant screen.

The condition for the first minimum (the first dark fringe) in this diffraction pattern is given by the elegant equation:
$$d \sin	heta = \lambda$$
where $	heta$ is the angular distance from the central maximum to the first minimum, and $\lambda$ is the wavelength of the light.

Our goal is to find the fractional error in the measurement of the wavelength, which is represented by $\frac{\Delta\lambda}{\lambda}$.

### The Mathematics of Error

To extract the fractional error from our equation, we use a powerful mathematical tool: **logarithmic differentiation**. By taking the natural logarithm of both sides, we transform the product into a sum:
$$\ln \lambda = \ln d + \ln(\sin	heta)$$

Now, we differentiate this equation. The derivative of $\ln x$ is $\frac{1}{x} dx$. Applying this, we get:
$$\frac{d\lambda}{\lambda} = \frac{dd}{d} + \frac{\cos	heta}{\sin	heta} d	heta$$

In the context of error analysis, we replace the differentials with absolute errors ($\Delta$) and ensure all terms are added to find the maximum possible error:
$$\frac{\Delta\lambda}{\lambda} = \frac{\Delta d}{d} + \cot	heta \cdot \Delta	heta$$

### The Trap: Degrees vs. Radians

Here is where many students stumble. The problem gives us the error in the angle as $\Delta	heta = 40'$. It is incredibly tempting to just plug this number straight into the equation. But beware!

**In calculus, whenever an angle stands alone (outside of a trigonometric function like sine or cosine), it MUST be in radians.**

Let's carefully convert $40'$ (arc minutes) into radians. First, we convert it to degrees by dividing by 60:
$$\Delta	heta = \left(\frac{40}{60}ight)^\circ = \left(\frac{2}{3}ight)^\circ$$

Next, we multiply by $\frac{\pi}{180}$ to convert degrees to radians:
$$\Delta	heta = \frac{2}{3} 	imes \frac{\pi}{180} = \frac{\pi}{270} 	ext{ rad}$$

### The Final Calculation

Now we are ready to bring it all together. We know:
- $d = 0.016 	ext{ mm}$
- $\Delta d = 0.002 	ext{ mm}$
- $	heta = 2^\circ$
- $\sin(2^\circ) = 0.035$

Let's substitute these into our error equation. We can rewrite $\cot	heta$ as $\frac{\sqrt{1 - \sin^2	heta}}{\sin	heta}$:
$$\frac{\Delta\lambda}{\lambda} = \frac{0.002}{0.016} + \frac{\sqrt{1 - (0.035)^2}}{0.035} 	imes \frac{\pi}{270}$$

The first term simplifies beautifully:
$$\frac{0.002}{0.016} = \frac{1}{8} = 0.125$$

For the second term, since $0.035$ is very small, its square is negligible, making the numerator $\sqrt{1 - (0.035)^2} \approx 1$. 
$$\frac{1}{0.035} 	imes \frac{\pi}{270} \approx 28.57 	imes 0.0116 \approx 0.332$$

Adding these two components together gives us our final answer:
$$\frac{\Delta\lambda}{\lambda} = 0.125 + 0.332 = 0.457$$

Rounding to two decimal places, the fractional error in the measurement of the wavelength is **0.46**. This problem is a fantastic reminder that in physics, the devil is always in the details—especially when it comes to units!

Enter Numerical Value:

Visualized Solution

$d sin θ = λ$

$\frac{d λ}{λ} = \frac{dd}{d} + cot θ d θ$

$Given Parameters$

$Δ θ in Radians$

$Substituting Values$

$Final Calculation$

The Sigma Insight: Diffraction due to Single Slit

The Setup

The Mathematics of Error

The Trap

The Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dsinθ=λ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λdλ​=ddd​+cotθdθ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Given Parameters

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δθ in Radians

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation