A grating \(1.600 \mathrm{cm}\) wide has exactly 12000 slits. The grating is used to resolve two nearly equal wavelengths in a light source: \(\lambda_{\mathrm{a}}=440.000 \mathrm{nm}\) and $\lambda_{\mathrm{b}}=440.936 \mathrm{nm}$ (a) How many orders of the lines can be seen with the grating? (b) What is the angular separation \(\theta_{b}-\theta_{a}\) between the lines in each order? (c) Which order best resolves the two lines? Explain.

To find the grating spacing, we can use the formula, \(d = \frac{W}{N}\), where W is the width of the grating and N is the number of slits. In this case, W = 1.600 cm and N = 12000. Therefore, $$d = \frac{1.600 \text{ cm}}{12000} = \frac{1.600 \times 10^{-2} \text{ m}}{12000} = 1.333 \times 10^{-6} \text{ m}$$. This is the distance between adjacent slits.

The grating equation is given by $$m\lambda = d\sin \theta $$ Since \(\theta\) can have a maximum value of 90°, we have \(\sin \theta_\text{max} = 1\). Thus, the maximum order 'm' for both wavelengths can be obtained by dividing both sides of the equation by the corresponding wavelength and then rounding down to the nearest integer: $$m_\text{max} = \left \lfloor \frac{d}{\lambda} \right \rfloor $$ For \(\lambda_\text{a}\), $$m_\text{max,a} = \left \lfloor \frac{1.333\times 10^{-6} \text{ m}}{440.000 \times 10^{-9} \text{ m}} \right \rfloor = 3$$, and for \(\lambda_\text{b}\), $$m_\text{max,b} = \left \lfloor \frac{1.333\times 10^{-6} \text{ m}}{440.936 \times 10^{-9} \text{ m}} \right \rfloor = 3$$ So, the maximum order that can be seen for both wavelengths is m = 3.

To find the angular separation, we can use the grating equation for both wavelengths and take the difference: $$|\theta_\text{b} - \theta_\text{a}| = \arcsin \left( \frac{m \lambda_\text{b}}{d} \right) - \arcsin \left( \frac{m \lambda_\text{a}}{d} \right) $$ For m = 1, $$|\theta_\text{b} - \theta_\text{a}|_1 = \arcsin \left( \frac{1(440.936 \times 10^{-9}\text{ m})}{1.333 \times 10^{-6} \text{ m}} \right) - \arcsin \left( \frac{1(440.000 \times 10^{-9}\text{ m})}{1.333 \times 10^{-6} \text{ m}} \right) \approx 0.0206°$$ For m = 2, $$|\theta_\text{b} - \theta_\text{a}|_2 = \arcsin \left( \frac{2(440.936 \times 10^{-9}\text{ m})}{1.333 \times 10^{-6} \text{ m}} \right) - \arcsin \left( \frac{2(440.000 \times 10^{-9}\text{ m})}{1.333 \times 10^{-6} \text{ m}} \right) \approx 0.0413°$$ For m = 3, $$|\theta_\text{b} - \theta_\text{a}|_3 = \arcsin \left( \frac{3(440.936 \times 10^{-9}\text{ m})}{1.333 \times 10^{-6} \text{ m}} \right) - \arcsin \left( \frac{3(440.000 \times 10^{-9}\text{ m})}{1.333 \times 10^{-6} \text{ m}} \right) \approx 0.0621°$$ The angular separations between the lines in each order are approximately 0.0206°, 0.0413°, and 0.0621°.

According to the Rayleigh criterion, if the angular separation between the peaks of the two lines meets or exceeds the criterion, the lines can be considered resolved. The Rayleigh criterion for the grating is given by: $$R_\text{min} = \frac{\lambda}{\Delta \lambda} = \frac{m}{\Delta m} $$, where $$\Delta \lambda = \lambda_\text{b} - \lambda_\text{a} = 440.936 \times 10^{-9} \text{ m} - 440.000 \times 10^{-9} \text{ m} = 0.936 \times 10^{-9} \text{ m}$$ For m = 1, $$R_\text{min} = \frac{1}{\Delta m} = \frac{440 \times 10^{-9} \text{ m}}{0.936 \times 10^{-9} \text{ m}} \approx 470$$ For m = 2, $$R_\text{min} = \frac{2}{\Delta m} = \frac{880 \times 10^{-9} \text{ m}}{0.936 \times 10^{-9} \text{ m}} \approx 940$$ For m = 3, $$R_\text{min} = \frac{3}{\Delta m} = \frac{1320 \times 10^{-9} \text{ m}}{0.936 \times 10^{-9} \text{ m}} \approx 1410$$ Based on the calculated values of R_min for each order, the grating should be able to resolve the two wavelengths reasonably well. However, the higher the order (m), the better the resolving power. In our case, the 3rd order (m = 3) best resolves the two lines because it has the largest angular separation (0.0621°) and the highest resolving power (R_min ≈ 1410).