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(a) How many rulings must a \(4.25 \mathrm{~cm}\) wide diffraction grating have to resolve the wavelengths \(415.496 \mathrm{~nm}\) and \(415.487 \mathrm{~nm}\) in the second order? (b) At what angle are the second-order maxima found?

Short Answer

Expert verified
(a) The diffraction grating must have 23083 rulings. (b) The second-order maxima are found at an angle of approximately 26.76°.

Step by step solution

01

Understand the problem

We have a diffraction grating of width \(4.25 \text{ cm}\), and we need to resolve two very close wavelengths: \( \lambda_1 = 415.496 \text{ nm} \) and \( \lambda_2 = 415.487 \text{ nm} \) in the second order of diffraction. The problem asks for the number of rulings (or lines) per grating, and for the angle at which the second-order maxima occur.
02

Use the Rayleigh Criterion for Resolution

The Rayleigh criterion for resolving two wavelengths \( \lambda_1 \) and \( \lambda_2 \) is given by:\[\Delta \lambda = \lambda_2 - \lambda_1 = \frac{\lambda}{N} \cdot m\]where \( \Delta \lambda = 415.496 \text{ nm} - 415.487 \text{ nm} = 0.009 \text{ nm} \), \( N \) is the total number of rulings, \( m \) is the order of diffraction (which is 2 in this problem), and \( \lambda \) is the resolving power: `(lambda + lambda')/2.`The resolving power is \( R = \frac{\lambda}{\Delta \lambda} \ge m \cdot N \). Here, average \( \lambda \) is \( \frac{1}{2}(415.496 + 415.487) = 415.4915 \text{ nm}\). Substitute this into the equation to find \( N \):\[R = \frac{415.4915}{0.009} = 46165.72\]
03

Calculate the Number of Rulings

Rearrange the Rayleigh criterion to solve for \( N \):\[N = \frac{R}{m} \le \frac{46165.72}{2}\]Solving this, we get:\[N = 23082.86\]Since \( N \) must be a whole number, round this up to the nearest whole number, giving us \( N = 23083 \). This means the grating must have 23083 rulings.
04

Determine the Angle of Second-order Maxima

The diffraction grating equation is given by:\[d \cdot \sin \theta = m \cdot \lambda\]We solve for \( \theta \) where \( d \) is the distance between adjacent rulings, \( m = 2 \) is the order, and \( \lambda = 415.4915 \text{ nm} \):First find \( d \): since the grating width is \(4.25 \text{ cm} = 42500 \text{ nm}\), and there are 23083 rulings,\[d = \frac{42500 \text{ nm}}{23083} \approx 1.841 \text{ nm}\]Now, use the grating equation:\[1.841 \cdot \sin \theta = 2 \cdot 415.4915\]Simplify and solve for \( \theta \):\[\sin \theta = \frac{2 \times 415.4915}{1.841}\]\[\sin \theta \approx 0.451\]\[\theta = \arcsin(0.451) \approx 26.76^\circ\]

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Key Concepts

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

Rayleigh criterion
The Rayleigh Criterion is essential when dealing with the resolution of close spectral lines through a diffraction grating. It establishes the minimum resolvable wavelength difference for a diffraction grating, indicating when two light sources can barely be seen as separate. This criterion is mathematically expressed as:
  • \( \Delta \lambda = \frac{\lambda}{N} \cdot m \)
where \( \Delta \lambda \) is the difference between wavelengths, \( \lambda \) represents the average wavelength, \( N \) is the number of rulings on the grating, and \( m \) denotes the diffraction order. The criterion essentially shows that to resolve very close wavelengths effectively, a greater number of rulings is necessary. The practical implication is that for better resolution, the grating should have a sufficient number of lines to separate the spectral lines, even in high orders of diffraction.
resolving power
The resolving power of a diffraction grating is a measure of its ability to separate close wavelengths. This is crucial in applications such as spectroscopy, where precise measurements of wavelengths are needed. Resolving power \( R \) is calculated as:
  • \( R = \frac{\lambda}{\Delta \lambda} \ge m \cdot N \)
where \( \lambda \) is the average wavelength, \( \Delta \lambda \) is the smallest resolved wavelength difference, \( N \) is the number of lines, and \( m \) is the order of diffraction. High resolving power indicates a grating's capability to distinguish between wavelengths that are very close together. This is vital for detailed spectral analysis, helping scientists and engineers in tasks ranging from experimental physics to the development of optical instruments.
diffraction order
Diffraction order is a fundamental concept when analyzing how light interacts with a diffraction grating. It refers to the periodic occurrences of maxima observed as light diffracts through the grating. Named by the integer \( m \), where 1 represents the first order, 2 the second, and so on, the diffraction order affects how closely wavelengths can be spaced while still being resolved.
  • The equation for maxima is \( d \cdot \sin \theta = m \cdot \lambda \),
where \( d \) is the distance between grating lines, and \( m \) is the diffraction order. Higher orders require precise calculations of angle and wavelength since each rising order contains more separated maxima. Understanding diffraction orders is critical for utilizing gratings in different applications, controlling the resolution by choosing suitable order number \( m \) for the required analysis.
angle of diffraction
The angle of diffraction is the angle at which light rays emerge after passing through a diffraction grating, creating patterns referred to as diffraction maxima. It is intimately connected with the order of diffraction \( m \) and can be calculated with the grating equation:
  • \( d \cdot \sin \theta = m \cdot \lambda \)
where \( d \) is the distance between adjacent grating lines and \( \lambda \) is the wavelength. Solving for \( \theta \) gives us the specific angle at which particular wavelengths appear for a given order. Correctly determining the angle of diffraction is crucial for applications like spectrometers, allowing them to identify precise wavelengths and their intensities. It aids researchers and engineers in designing optical systems and interpreting light spectra for accurate measurements and analyses.

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Most popular questions from this chapter

A slit \(1.00 \mathrm{~mm}\) wide is illuminated by light of wavelength \(650 \mathrm{~nm}\). We see a diffraction pattern on a screen \(3.00 \mathrm{~m}\) away. What is the distance between the first and third diffraction minima on the same side of the central diffraction maximum?

a) How many bright fringes appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern if \(\lambda=550 \mathrm{~nm}, d=0.180 \mathrm{~mm}\), and \(a=30.0 \mu \mathrm{m}\) ? (b) What is the ratio of the intensity of the third bright fringe to the intensity of the central fringe?

What must be the ratio of the slit width to the wavelength for a single slit to have the first diffraction minimum at \(\theta=30^{\circ}\) ?

A plane wave of wavelength \(420 \mathrm{~nm}\) is incident on a slit with a width of \(a=0.60 \mathrm{~mm}\). A thin converging lens of focal length \(+70 \mathrm{~cm}\) is placed between the slit and a viewing screen and focuses the light on the screen. (a) How far is the screen from the lens? (b) What is the distance on the screen from the center of the diffraction pattern to the first minimum?

A diffraction grating having 180 lines/mm is illuminated with a light signal containing only two wavelengths, \(A_{1}=400 \mathrm{~nm}\) and \(\lambda_{2}=600 \mathrm{~nm}\). The signal is incident perpendicularly on the grating. (a) What is the angular separation between the secondorder maxima of these two wavelengths? (b) What is the smallest angle at which two of the resulting maxima are superimposed? (c) What is the highest order for which maxima for both wavelengths are present in the diffraction pattern?

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