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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

Expert verified
The grating must have at least 2755 slits per centimeter.

Step by step solution

01

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.
02

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.
03

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 \( 587.9782 = N(0.178) \).
04

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 \( N = 3305 \).
05

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.
06

Conclusion

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Why is visible light, which has much longer wavelengths than x rays do, used for Bragg reflection experiments on colloidal crystals? (a) The microspheres are suspended in a liquid, and it is more difficult for x rays to penetrate liquid than it is for visible light. (b) The irregular spacing of the microspheres allows the longerwavelength visible light to produce more destructive interference than can x rays. (c) The microspheres are much larger than atoms in a crystalline solid, and in order to get interference maxima at reasonably large angles, the wavelength must be much longer than the size of the individual scatterers. (d) The microspheres are spaced more widely than atoms in a crystalline solid, and in order to get interference maxima at reasonably large angles, the wavelength must be comparable to the spacing between scattering planes.

Different isotopes of the same element emit light at slightly different wavelengths. A wavelength in the emission spectrum of a hydrogen atom is 656.45 nm; for deuterium, the corresponding wavelength is 656.27 nm. (a) What minimum number of slits is required to resolve these two wavelengths in second order? (b) If the grating has 500.00 slits/mm, find the angles and angular separation of these two wavelengths in the second order.

Your physics study partner tells you that the width of the central bright band in a single-slit diffraction pattern is inversely proportional to the width of the slit. This means that the width of the central maximum increases when the width of the slit decreases. The claim seems counterintuitive to you, so you make measurements to test it. You shine monochromatic laser light with wavelength \(\lambda\) onto a very narrow slit of width \(a\) and measure the width \(w\) of the central maximum in the diffraction pattern that is produced on a screen 1.50 m from the slit. (By "width," you mean the distance on the screen between the two minima on either side of the central maximum.) Your measurements are given in the table. (a) If \(w\) is inversely proportional to \(a\), then the product \(aw\) is constant, independent of \(a\). For the data in the table, graph \(aw\) versus \(a\). Explain why \(aw\) is not constant for smaller values of \(a\). (b) Use your graph in part (a) to calculate the wavelength \(\lambda\) of the laser light. (c) What is the angular position of the first minimum in the diffraction pattern for (i) \(a\) = 0.78 \(\mu\)m and (ii) \(a\) = 15.60 \(\mu\)m?

When laser light of wavelength 632.8 nm passes through a diffraction grating, the first bright spots occur at \(\pm\)17.8\(^\circ\) from the central maximum. (a) What is the line density (in lines/cm) of this grating? (b) How many additional bright spots are there beyond the first bright spots, and at what angles do they occur?

An interference pattern is produced by light of wavelength 580 nm from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.530 mm. (a) If the slits are very narrow, what would be the angular positions of the first-order and second-order, two-slit interference maxima? (b) Let the slits have width 0.320 mm. In terms of the intensity \(I_0\) at the center of the central maximum, what is the intensity at each of the angular positions in part (a)?

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