In a single slit diffraction experiment, a slit of width (0.016 ±...

For a single slit diffraction, the condition for the first minimum is given by:

$a \sin \theta = \lambda$

Taking the natural logarithm on both sides, we get:

$\ln \lambda = \ln a + \ln(\sin \theta)$

Differentiating to find the maximum fractional error:

$\dfrac{\Delta \lambda}{\lambda} = \dfrac{\Delta a}{a} + \cot \theta \Delta \theta$

Since $\theta = 2^\circ$ is very small, $\cos \theta \approx 1$, which gives $\cot \theta \approx \dfrac{1}{\sin \theta}$.

Given $\sin(2^\circ) = 0.035$, we can approximate the angle in radians as $\theta \approx \sin \theta = 0.035$ rad.

The error in the angle is $\Delta \theta = 40' = \left(\dfrac{40}{60}\right)^\circ = \left(\dfrac{2}{3}\right)^\circ$.

Converting $\Delta \theta$ into radians:

$\Delta \theta = \dfrac{2}{3} \times \left(\dfrac{0.035}{2}\right) = \dfrac{0.035}{3}$ rad

The fractional error in the angular term is:

$\cot \theta \Delta \theta \approx \dfrac{\Delta \theta}{\sin \theta} = \dfrac{\dfrac{0.035}{3}}{0.035} = \dfrac{1}{3}$

The fractional error in the slit width is:

$\dfrac{\Delta a}{a} = \dfrac{0.002}{0.016} = \dfrac{1}{8}$

Substituting these values into the error equation:

$\dfrac{\Delta \lambda}{\lambda} = \dfrac{1}{8} + \dfrac{1}{3} = \dfrac{3 + 8}{24} = \dfrac{11}{24}$

Answer: $0.46$