1. **Problem statement:** (a) Calculate the angular width of the second-order maximum for a diffraction grating with 400 lines/mm illuminated by light wavelengths from $4.00 \times 10^{-7}$ m to $7.00 \times 10^{-7}$ m. (b) Determine the largest diffraction order for which all wavelengths are present. 2. **Relevant formula:** The diffraction grating equation for maxima is: $$d \sin \theta = m \lambda$$ where $d$ is the grating spacing, $m$ is the order number, and $\lambda$ is the wavelength. 3. **Calculate grating spacing $d$:** Given 400 lines/mm, convert to meters: $$d = \frac{1}{400 \times 10^{3}} = 2.5 \times 10^{-6} \text{ m}$$ 4. **(a) Angular width of second-order maximum:** The angular width $\Delta \theta$ is the difference between angles for the longest and shortest wavelengths at order $m=2$: $$\Delta \theta = \theta_{max} - \theta_{min}$$ where $$\sin \theta_{max} = \frac{2 \times 7.00 \times 10^{-7}}{2.5 \times 10^{-6}} = 0.56$$ $$\sin \theta_{min} = \frac{2 \times 4.00 \times 10^{-7}}{2.5 \times 10^{-6}} = 0.32$$ Calculate angles: $$\theta_{max} = \arcsin(0.56) \approx 34.0^\circ$$ $$\theta_{min} = \arcsin(0.32) \approx 18.7^\circ$$ Therefore, $$\Delta \theta = 34.0^\circ - 18.7^\circ = 15.3^\circ$$ 5. **(b) Largest order $m_{max}$ for all wavelengths present:** The maximum order is limited by the longest wavelength: $$m_{max} = \left\lfloor \frac{d}{\lambda_{max}} \right\rfloor$$ Calculate: $$m_{max} = \left\lfloor \frac{2.5 \times 10^{-6}}{7.00 \times 10^{-7}} \right\rfloor = \left\lfloor 3.57 \right\rfloor = 3$$ **Final answers:** (a) Angular width of second-order maximum is approximately $15.3^\circ$. (b) The largest order for which all wavelengths are present is 3.