(a) If a single slit produces a first minimum at \(14.5^{\circ}\) at what angle is the second-order minimum? (b) What is the angle of the third-order minimum? (c) Is there a fourthorder minimum? (d) Use your answers to illustrate how the angular width of the central maximum is about twice the angular width of the next maximum (which is the angle between the first and second minima).

We are given that the first minimum occurs at an angle of \(14.5^\circ\). We can use the formula for the angle of minimum in a single slit diffraction: \[ \sin{\theta} = \frac{m \lambda}{a} \] We know that \[\sin{14.5^\circ} = \frac{1 \lambda}{a} \] (for \(m = 1\)) Now, we'll find the second-order minimum angle, using the same formula with \(m = 2\): \[\sin{\theta_2} = \frac{2 \lambda}{a} \] Since the slit width, \(a\), and the wavelength, \(\lambda\), are the same for both angles, we can divide the two equations: \[\frac{\sin{\theta_2}}{\sin{14.5^\circ}} = \frac{2 \lambda / a}{\lambda / a} \] Simplifying and solving for \(\theta_2\): \[\theta_2 = \arcsin{(2 \sin{14.5^\circ})}\] Calculating the angle: \[\theta_2 \approx 30.1^{\circ}\] So, the second-order minimum is at an angle of approximately \(30.1^\circ\).

We'll find the third-order minimum angle in a similar manner, using the same formula with \(m = 3\): \[\sin{\theta_3} = \frac{3 \lambda}{a} \] Dividing the equations (for \(m=3\) and \(m=1\)): \[\frac{\sin{\theta_3}}{\sin{14.5^\circ}} = \frac{3 \lambda / a}{\lambda / a} \] Solving for \(\theta_3\): \[\theta_3 = \arcsin{(3 \sin{14.5^\circ})}\] Calculating the angle: \[\theta_3 \approx 48.5^{\circ}\] So, the third-order minimum is at an angle of approximately \(48.5^\circ\).

We can check if there is a fourth-order minimum by calculating the angle for \(m = 4\): \[\sin{\theta_4} = \frac{4 \lambda}{a} \] Dividing the equations (for \(m=4\) and \(m=1\)): \[\frac{\sin{\theta_4}}{\sin{14.5^\circ}} = \frac{4 \lambda / a}{\lambda / a} \] Solving for \(\theta_4\): \[\theta_4 = \arcsin{(4 \sin{14.5^\circ})}\] However, this calculation results in a value greater than 1 for \(\sin{\theta_4}\), which is not possible. Therefore, there is no fourth-order minimum.

Finally, we'll compare the angular widths of the central maximum and the angular width of the next maximum. The central maximum extends between the minima on either side of the \(0^\circ\) angle, so its angular width is: \[\Delta \theta_{central} = 2 \cdot 14.5^{\circ} = 29^{\circ} \] The next maximum is between the first and second minima, so its angular width is: \[\Delta \theta_{next} = 30.1^{\circ} - 14.5^{\circ} = 15.6^{\circ} \] Comparing the two: \[\frac{\Delta \theta_{central}}{\Delta \theta_{next}} = \frac{29^{\circ}}{15.6^{\circ}} \approx 1.86 \] The ratio of the angular width of the central maximum to the next maximum is approximately 1.86, which is close to 2 as stated in the question. This validates the observation that the angular width of the central maximum is about twice the angular width of the next maximum.