A particular transparent diffraction grating is ruled with 5,500 lines per centimeter, and \(2.5 \mathrm{~cm}\) of it is used in a simple spectrometer having a source slit, collimating lens, and telescope. At what angles are the sodium D lines \((\lambda=589.0\) and \(589.6 \mathrm{~nm})\) found if the grating is illuminated at normal incidence? What is the dispersion in minutes of arc per nanometer and the theoretical resolution at each of these angular positions?

Now, we should convert the distance to meters for easier calculations: $$d = \frac{1}{5500} \mathrm{~cm} \times \frac{1\mathrm{~m}}{100\mathrm{~cm}} = 1.818\times10^{-5} \mathrm{~m}$$ #Step 2: Solve for Angles of Diffraction for Sodium D Lines# Using the grating equation, we can solve for the angles of diffraction for the sodium D lines. Let's start with \(\lambda_1 = 589.0 \mathrm{~nm}\): $$n\lambda_1 = d \sin(\theta_1)$$ $$\theta_1 = \sin^{-1}\left(\frac{n\lambda_1}{d}\right)$$ We must check for each order of diffraction (n) if it produces a valid angle between 0° and 90°. Let's calculate the angles for both D lines and the first two orders of diffraction (n = 1 and n = 2): - For \(\lambda_1 = 589.0 \mathrm{~nm}\) and n = 1: $$\theta_1 = \sin^{-1}\left(\frac{1\cdot(589.0 \times 10^{-9} \mathrm{~m})}{1.818\times10^{-5} \mathrm{~m}}\right) \approx 18.98°$$ - For \(\lambda_1 = 589.0 \mathrm{~nm}\) and n = 2: $$\theta_1 = \sin^{-1}\left(\frac{2\cdot(589.0 \times 10^{-9} \mathrm{~m})}{1.818\times10^{-5} \mathrm{~m}}\right)$$ This calculation results in a value greater than 1 inside the sin inverse function, which means there is no second-order diffraction for \(\lambda_1\). - For \(\lambda_2 = 589.6 \mathrm{~nm}\) and n = 1: $$\theta_2 = \sin^{-1}\left(\frac{1\cdot(589.6 \times 10^{-9} \mathrm{~m})}{1.818\times10^{-5} \mathrm{~m}}\right) \approx 18.99°$$ - For \(\lambda_2 = 589.6 \mathrm{~nm}\) and n = 2: Again, there is no second-order diffraction for \(\lambda_2\). So, the angles of diffraction for the first-order sodium D lines are approximately 18.98° and 18.99°. #Step 3: Calculate Angular Dispersion# Now, we can calculate the angular dispersion of the grating using the derivative of the grating equation. For simplicity, let's take the first order (n = 1): $$\frac{d\theta}{d\lambda} = \frac{1}{d\cos(\theta)}$$ Using the angle calculated for \(\lambda_1 = 589.0 \mathrm{~nm}\): $$\frac{d\theta}{d\lambda} = \frac{1}{(1.818\times10^{-5} \mathrm{~m})\cos(18.98°)} \approx 4.806 \times 10^6 \mathrm{~m}^{-1}$$ The angular dispersion for the first-order sodium D lines is approximately \(4.806 \times 10^6 \mathrm{~m}^{-1}\). #Step 4: Calculate Theoretical Resolution# Finally, we can calculate the theoretical resolution of the grating. First, let's find the total number of rulings (N) in the grating. Assuming that the grating is 1 cm wide: $$N = 1\mathrm{~cm} \times 5500\frac{\text{lines}}{\mathrm{cm}} = 5500$$ Using the previously mentioned equation for resolution and the first order (n = 1): $$R = nN = 1\cdot 5500 = 5500$$ The theoretical resolution of the grating for the first-order sodium D lines is 5,500.