Using the expression 2d sin heta = \lambda, one calculates the values of d by measuring the corres

Q: Why do we use logarithmic differentiation here?

Logarithmic differentiation turns multiplication and division into addition and subtraction, making it much easier to apply the rules of error analysis.

Q: Why is the error in $\theta$ considered constant?

The problem explicitly states that the error in measuring $\theta$ ($\Delta\theta$) is constant, which is typical for a given measuring instrument like a spectrometer.

Q: What is the difference between absolute and fractional error?

Absolute error ($\Delta d$) is the actual amount by which the measurement might be off. Fractional error ($\frac{\Delta d}{d}$) is the ratio of the absolute error to the measured value, giving a sense of the relative precision.

Q: Why does $\cot\theta$ decrease as $\theta$ increases?

In the first quadrant (0 to 90 degrees), $\cos\theta$ decreases from 1 to 0, while $\sin\theta$ increases from 0 to 1. Since $\cot\theta = \frac{\cos\theta}{\sin\theta}$, the ratio of a decreasing numerator to an increasing denominator must decrease.

Q: Could the absolute error increase while the fractional error decreases?

Yes, it's mathematically possible if the measured value $d$ increases faster than the fractional error decreases. However, in this specific problem, we proved that both errors decrease.

Question

Using the expression $2d \sin 	heta = \lambda$, one calculates the values of $d$ by measuring the corresponding angles $	heta$ in the range $0^{\circ}$ to $90^{\circ}$. The wavelength $\lambda$ is exactly known and the error in $	heta$ is constant for all values of $	heta$. As $	heta$ increases from $0^{\circ}$

Options:

(A) the absolute error in $d$ remains constant

(B) the absolute error in $d$ increases

(C) the fractional error in $d$ remains constant

(D) the fractional error in $d$ decreases

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The problem asks us to analyze how the errors in calculating the slit separation $d$ change as the measured angle $	heta$ increases. We are given the relation $2d \sin 	heta = \lambda$, where the wavelength $\lambda$ is a known constant, and the error in measuring the angle, $\Delta	heta$, is also constant.

Our first step is to express the quantity we are calculating, $d$, in terms of the measured quantity, $	heta$. Rearranging the given equation, we get:
$$d = \frac{\lambda}{2\sin	heta}$$

### The Master Equation for Errors

To find the fractional error $\frac{\Delta d}{d}$, the most elegant method is logarithmic differentiation. By taking the natural logarithm of both sides, we convert the division into subtraction, which is much easier to differentiate.
$$\ln d = \ln \lambda - \ln 2 - \ln \sin	heta$$

Now, we differentiate this expression. Remember that $\lambda$ and $2$ are constants, so their derivatives are zero.
$$\frac{\Delta d}{d} = 0 - 0 - \frac{1}{\sin	heta} \cdot \cos	heta \cdot \Delta	heta$$
$$\frac{\Delta d}{d} = -\cot	heta \cdot \Delta	heta$$

When dealing with errors, we are interested in the maximum possible error, so we take the absolute value (magnitude):
$$\left|\frac{\Delta d}{d}ight| = \cot	heta \cdot \Delta	heta$$

### Final Calculation and Conclusion

We are given that $\Delta	heta$ is constant. Now we need to see how the fractional error behaves as $	heta$ increases from $0^{\circ}$ to $90^{\circ}$. In the first quadrant, as the angle increases, the value of the cotangent function ($\cot	heta$) strictly decreases. Since the fractional error is directly proportional to $\cot	heta$, it must also decrease.

Let's also quickly check the absolute error, $\Delta d$, just to be thorough. We can find it by multiplying the fractional error by $d$:
$$\Delta d = d \cdot \cot	heta \cdot \Delta	heta$$
Substituting $d = \frac{\lambda}{2\sin	heta}$, we get:
$$\Delta d = \left(\frac{\lambda}{2\sin	heta}ight) \left(\frac{\cos	heta}{\sin	heta}ight) \Delta	heta = \frac{\lambda \cos	heta}{2\sin^2	heta} \Delta	heta$$

As $	heta$ increases from $0^{\circ}$ to $90^{\circ}$, the numerator $\cos	heta$ decreases, while the denominator $\sin^2	heta$ increases. A decreasing numerator divided by an increasing denominator means the entire fraction decreases. Thus, the absolute error $\Delta d$ also decreases.

Looking at the given options, the only correct statement is that **the fractional error in $d$ decreases**.

Select Answer:

Visualized Solution

$d = \frac{λ}{2 s i n θ}$

$Logarithmic Differentiation$

$Fractional Error Analysis$

$Absolute Error Analysis$

$Conclusion$

The Sigma Insight: Errors in Measurement

Analyzing the Setup

The Master Equation for Errors

Final Calculation and Conclusion

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