In a single slit diffraction experiment, a slit of width (0.016 ± 0.002) 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° ± 40'). The value of the fractional error in the measurement of wavelength is: Given$: \sin(2°) = 0.035$

Step 1: Identify the given parameters.

The slit width $a$ and the angular position of the first minimum $\theta$ are given with their respective absolute errors.

$$a = 0.016\ \text{mm}$$ $$\Delta a = 0.002\ \text{mm}$$ $$\theta = 2^\circ$$ $$\Delta\theta = 40'$$ $$\sin 2^\circ = 0.035$$

Step 2: State the condition for the first minimum in single slit diffraction.

The relation between the slit width, diffraction angle, and wavelength is given by the diffraction formula.

$$a\sin\theta = \lambda$$

Step 3: Derive the expression for the fractional error in wavelength.

Taking the natural logarithm on both sides and differentiating gives the relative error relation.

$$\ln\lambda = \ln a + \ln(\sin\theta)$$ $$\frac{\Delta\lambda}{\lambda} = \frac{\Delta a}{a} + \frac{\cos\theta}{\sin\theta}\Delta\theta$$ $$\frac{\Delta\lambda}{\lambda} = \frac{\Delta a}{a} + \cot\theta\,\Delta\theta$$

Step 4: Calculate the individual terms.

Convert the angular error $\Delta\theta$ to radians and compute the fractional error of the slit width.

$$\frac{\Delta a}{a} = \frac{0.002}{0.016} = 0.125$$ $$\Delta\theta = \frac{40}{60} \times \frac{\pi}{180}\ \text{rad} \approx 0.0116\ \text{rad}$$ $$\cot 2^\circ = \frac{\sqrt{1 - \sin^2 2^\circ}}{\sin 2^\circ}$$ $$\cot 2^\circ = \frac{\sqrt{1 - (0.035)^2}}{0.035} \approx 28.6$$

Step 5: Calculate the total fractional error.

Substitute the calculated values back into the fractional error formula.

$$\frac{\Delta\lambda}{\lambda} = 0.125 + 28.6 \times 0.0116$$ $$\frac{\Delta\lambda}{\lambda} = 0.125 + 0.332$$ $$\frac{\Delta\lambda}{\lambda} = 0.457 \approx 0.46$$

Answer: The fractional error in the measurement of wavelength is approximately $0.46$ (or $46\%$).