Question

A grating spectrometer is used to resolve wavelengths \(660.0 \mathrm{nm}\) and \(661.4 \mathrm{nm}\) in second order. (a) How many slits per centimeter must the grating have to produce both wavelengths in second order? (The answer is either a maximum or a minimum number of slits per centimeter.) (b) The minimum number of slits required to resolve two closely spaced lines is $N=\lambda /(m \Delta \lambda),\( where \)\lambda$ is the average of the two wavelengths, \(\Delta \lambda\) is the difference between the two wavelengths, and \(m\) is the order. What minimum number of slits must this grating have to resolve the lines in second order?

Short answer

In summary: a) The number of slits per centimeter, S, needed to resolve both 660.0 nm and 661.4 nm wavelengths in the second order lies within the range of minimum and maximum values, which are determined by the angles close to 0° and 90°. b) The minimum number of slits, N, required to resolve the two closely spaced lines in the second order is approximately 236 slits.

Step-by-step solution

(a) Using the grating equation

We need to solve for the number of slits per centimeter that can resolve both wavelengths in the second order. We will use the grating equation first: $$d \cdot \sin{θ} = m \cdot \lambda $$ Where d is the distance between the slits, m is the order, and λ is the wavelength. We are given the wavelengths and the order (second order): - \(λ_1 = 660.0 nm\) - \(λ_2 = 661.4 nm\) - \(m = 2\) We want to find the number of slits per centimeter, so let this number be S. Since d (distance between slits) will be equal to \(1/S\). Then we can rewrite the grating equation as: $$\frac{1}{S} = m \cdot \lambda/\sin{θ} $$ We have two grating equations (one per wavelength) with two unknowns (S and θ). It will be easier to work with one unknown. Notice that the angles are close to each other as we are talking about very close wavelengths. We can then use the small-angle approximation: $$\tan{θ} ≈ \sin{θ} $$ So, S can be found by $$S = \frac{\lambda}{\tan{θ}}$$ We then get the final equation for the first step: $$\frac{\lambda_1 - \lambda_2}{\lambda_1 \tan{θ_1} - \lambda_2 \tan{θ_2}} = \frac{m}{S} $$ Now, let's find S:

(a) Solving for S

Now we can use the given wavelengths and the order to solve for the number of slits per centimeter (S): $$\frac{660.0 \mathrm{nm} - 661.4 \mathrm{nm}}{660.0 \mathrm{nm} \tan{θ_1} - 661.4 \mathrm{nm} \tan{θ_2}} = \frac{2}{S} $$ As the wavelengths are close to each other, the difference between θ_1 and θ_2 will also be small, and we can consider the angles to be equal: θ_1 ≈ θ_2: $$\frac{660.0 \mathrm{nm} - 661.4 \mathrm{nm}}{(660.0 \mathrm{nm} - 661.4 \mathrm{nm}) \tan{θ}} = \frac{2}{S} $$ Now, we get: $$ S = \frac{2}{\tan{θ}}$$ Since we want to find the minimum or maximum number of slits per centimeter, the angles will be either at min or max. To find the minimum S, we can use the maximum θ, which occurs when θ = 90°. In this case, S would be a minimum, and the angle will be very close to 90° but not exactly 90°: $$S_{min} = \frac{2}{\tan{(90°-\epsilon)}}$$ Similarly, to find the maximum S, we can use the minimum θ, which occurs when θ = 0°. In this case, the angle will be very close to 0°: $$S_{max} = \frac{2}{\tan{\epsilon}}$$ To find the range of values of S, we can say: $$S_{min} \leq S \leq S_{max}$$ Thus, the number of slits per centimeter will be in the range of minimum and maximum values allowed.

(b) Using the minimum resolving power formula

Now, we need to find the minimum number of slits required to resolve the two closely spaced lines using the given formula: $$N = \frac{\lambda}{m \Delta \lambda} $$ Where N is the minimum number of slits required, λ is the average of the two wavelengths, m is the order, and Δλ is the difference between the wavelengths. We are given these values: - \(λ_1 = 660.0 \mathrm{nm}\) - \(λ_2 = 661.4 \mathrm{nm}\) - \(m = 2\) First, let's find the average wavelength, λ: $$\lambda = \frac{660.0 \mathrm{nm} + 661.4 \mathrm{nm}}{2} = 660.7 \mathrm{nm}$$ Now, let's find the difference between the wavelengths, Δλ: $$\Delta \lambda = 661.4 \mathrm{nm} - 660.0 \mathrm{nm} = 1.4 \mathrm{nm} $$ Finally, we can substitute the values into the minimum resolving power formula: $$N = \frac{660.7 \mathrm{nm}}{2 \cdot 1.4 \mathrm{nm}} = \frac{660.7}{2.8} \approx 236$$ The grating must have at least 236 slits to resolve the lines in the second order.