Light with wavelength \(633 \mathrm{nm}\) is incident on a \(2.5-\mu \mathrm{m}\) -wide slit. Find the angular width of the central peak in the diffraction pattern, taken as the angular separation between the first minima.

Firstly, we need to identify the relevant parameters provided in the problem: the wavelength of the light, \(\lambda = 633 \mathrm{nm}\), and the width of the slit, \(b = 2.5 \mu\mathrm{m}\). These values need to be converted to the same units for use in the calculations; we'll use meters, so \(\lambda = 633 \times 10^{-9} \mathrm{m}\) and \(b = 2.5 \times 10^{-6} \mathrm{m}\). Note that the order of the minima, \(m\), is 1 since we are looking at the first minima.

Next, we use the formula for the angle of the first minimum in a single-slit diffraction pattern: \( \sin \theta = \frac{m\lambda}{b} \). Substituting the appropriate values: \( \sin \theta = \frac{1\times 633\times 10^{-9}}{2.5\times 10^{-6}} \). Solving this gives us \(\sin \theta = 0.2532\). But we are looking for the angle in degrees. To go from the sine of the angle to the angle itself, we use the inverse sine function, also known as arcsine. So, \(\theta = \arcsin(0.2532)\), which gives us \(\theta \approx 14.5^\circ\). However, this is the angle to the first minimum, and we are interested in the full angular width of the central peak, which goes from the first minimum on one side to the first minimum on the other side.

The full angular width of the central peak in the diffraction pattern is twice the angle to the first minimum, so we simply multiply our result from step 2 by 2: \(2\times 14.5 \approx 29^\circ\).