Question

The light from an iron arc includes many different wavelengths. Two of these are at \(\lambda\) = 587.9782 nm and \(\lambda\) = 587.8002 nm. You wish to resolve these spectral lines in first order using a grating 1.20 cm in length. What minimum number of slits per centimeter must the grating have?

Short answer

The grating must have at least 2755 slits per centimeter.

Step-by-step solution

Understand the Problem

We need to resolve two spectral lines using a diffraction grating. Resolving power (R) of a grating helps to distinguish between close wavelengths. The formula for resolving power is \( R = \frac{\lambda}{\Delta \lambda} = mN \), where \( m \) is the order of the spectrum, \( N \) is the total number of slits, \( \lambda \) is one of the wavelengths, and \( \Delta \lambda \) is the difference in wavelengths.

Calculate Wavelength Difference

Identify the two wavelengths given: \( \lambda_1 = 587.9782 \) nm and \( \lambda_2 = 587.8002 \) nm. Calculate their difference: \( \Delta \lambda = \lambda_1 - \lambda_2 = 587.9782 - 587.8002 = 0.178 \) nm.

Set Up the Resolving Power Equation

We want to resolve these lines in first order, so \( m = 1 \). Substitute into the resolving power formula: \( R = \frac{587.9782}{0.178} = mN \), which simplifies to \( 587.9782 = N(0.178) \).

Solve for Total Number of Slits (N)

Calculate \( N \), the total number of slits: \( N = \frac{0.178}{0.178} \approx 3304.37 \). To resolve the lines, \( N \) must be an integer, so \( N = 3305 \).

Calculate Slits per Centimeter

The grating is 1.20 cm in length. Calculate the minimum number of slits per centimeter (d): \( d = \frac{N}{1.20} = \frac{3305}{1.20} \approx 2754.17 \) slits/cm.

Conclusion

Round up the value of slits per centimeter to ensure resolution: The grating must have a minimum of 2755 slits/cm.

Key concepts

Resolving Power

The resolving power of a diffraction grating is a measure of its ability to separate closely spaced spectral lines, such as those produced by light passing through a slit. It is an essential concept in optics that can be understood through a simple formula:

• \( R = \frac{\lambda}{\Delta \lambda} = mN \)

In this equation, \( R \) represents the resolving power, \( \lambda \) is the wavelength of light, \( \Delta \lambda \) is the smallest difference in wavelength that can be resolved, \( m \) is the order of the spectrum, and \( N \) is the total number of slits within the grating.
Resolving power is crucial in applications such as spectroscopy, where distinguishing between different substances relies on measuring specific wavelengths of light. The greater the resolving power, the more precise the separation of wavelengths. In practical terms, a higher \( N \) or using higher orders \( m \) can help increase the resolving power of the grating.

Wavelength Difference

The wavelength difference, symbolized as \( \Delta \lambda \), is the distinction in measurements between two close wavelengths. In the context of diffraction gratings and the exercise at hand, identifying \( \Delta \lambda \) is key to solving the problem of spectral line resolution.
For the given problem, the wavelengths are \( \lambda_1 = 587.9782 \) nm and \( \lambda_2 = 587.8002 \) nm. Calculating the difference involves subtracting these values:

• \( \Delta \lambda = \lambda_1 - \lambda_2 = 0.178 \) nm

This small difference exemplifies how closely spaced spectral lines can be and why a precise resolving power is crucial for identification. When this value is plugged into the resolving power formula, it helps determine whether the grating can effectively distinguish between these two given wavelengths.

Spectral Lines

Spectral lines are unique lines that represent specific wavelengths of light emitted or absorbed by substances. They act like optical fingerprints for different elements, helping scientists and engineers identify the composition of an object.
These lines originate from electron transitions within atoms or molecules. When electrons jump between energy levels, they emit or absorb light with characteristic wavelengths.
In the given exercise, the task is to resolve two spectral lines with wavelengths of \( 587.9782 \) nm and \( 587.8002 \) nm. Due to their closeness in wavelength, high resolving power is required from the diffraction grating. Accurately resolving these lines allows us to differentiate between similar elements or compounds, providing valuable insights in areas such as astronomy and chemical analysis.

Slits Per Centimeter

The concept of slits per centimeter describes how many slits are present within a single centimeter length of a diffraction grating. This measurement is crucial as it directly affects the grating's resolving power. The more slits packed into a given length, the higher its ability to separate close spectral lines.
For instance, in the original exercise, resolving power determines the number of slits necessary to discern two close wavelengths. The total number of slits \( N \) calculated was around 3305, and the grating had a length of 1.20 cm. Thus, finding the slits per centimeter involves:

• \( d = \frac{N}{1.20} \approx 2754.17 \) slits/cm

To ensure resolution, we round up to 2755 slits/cm. This high density of slits maximizes the grating's resolving capability, making it suitable for tasks like resolving closely spaced spectral lines.