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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

Laser light of wavelength \(500.0 \mathrm{nm}\) illuminates two identical slits, producing an interference pattern on a screen \(90.0 \mathrm{~cm}\) from the slits. The bright bands are \(1.00 \mathrm{~cm}\) apart, and the third bright bands on either side of the central maximum are missing in the pattern. Find the width and the separation of the two slits.

When the light is passed through the bottom of the sample container, the interference maximum is observed to be at 41\(^\circ\); when it is passed through the top, the corresponding maximum is at 37\(^\circ\). What is the best explanation for this observation? (a) The microspheres are more tightly packed at the bottom, because they tend to settle in the suspension. (b) The microspheres aremore tightly packed at the top, because they tend to float to the top of the suspension. (c) The increased pressure at the bottom makes the microspheres smaller there. (d) The maximum at the bottom corresponds to \(m\) = 2, whereas the maximum at the top corresponds to \(m\) = 1.

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?

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.

It has been proposed to use an array of infrared telescopes spread over thousands of kilometers of space to observe planets orbiting other stars. Consider such an array that has an effective diameter of 6000 km and observes infrared radiation at a wavelength of 10 \(\mu\)m. If it is used to observe a planet orbiting the star 70 Virginis, which is 59 light-years from our solar system, what is the size of the smallest details that the array might resolve on the planet? How does this compare to the diameter of the planet, which is assumed to be similar to that of Jupiter (1.40 \(\times\) 10\(^{5}\) km)? (Although the planet of 70 Virginis is thought to be at least 6.6 times more massive than Jupiter, its radius is probably not too different from that of Jupiter. Such large planets are thought to be composed primarily of gases, not rocky material, and hence can be greatly compressed by the mutual gravitational attraction of different parts of the planet.)

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