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Chapter 10: Problem 64

A grating has a total width of \(10.0 \mathrm{cm}\) and contains 600 lines/ \(\mathrm{mm} .\) What is its resolving power in the second-order spectrum? At a mean wavelength of \(540 \mathrm{nm}\), what wavelength difference can it resolve?

Short Answer

Expert verified
Resolving power is 120,000, and it can resolve a wavelength difference of 4.5 pm.

Step by step solution

01

Calculate Total Number of Lines

First, convert the total width of the grating into millimeters: \(10.0 \text{ cm} = 100 \text{ mm}\). Since there are 600 lines per millimeter, multiply the total width by the number of lines per mm: \(N = 100 \times 600 = 60,000\). So the grating contains 60,000 lines.
02

Use Resolving Power Formula

The resolving power \(R\) of a diffraction grating is given by the formula \(R = nN\), where \(n\) is the order of the spectrum and \(N\) is the number of lines. In the second-order spectrum, \(n = 2\). Therefore, the resolving power is \(R = 2 \times 60,000 = 120,000\).
03

Calculate Minimum Wavelength Difference

The resolving power \(R\) is also given by \(R = \frac{\lambda}{\Delta \lambda}\) where \(\Delta \lambda\) is the smallest wavelength difference that can be resolved and \(\lambda\) is the mean wavelength. Rearrange the formula to find \(\Delta \lambda = \frac{\lambda}{R}\). Substitute the given mean wavelength \(\lambda = 540\, \text{nm} = 540 \times 10^{-9} \text{m}\) and \(R = 120,000\) to find \(\Delta \lambda = \frac{540 \times 10^{-9}}{120,000} = 4.5 \times 10^{-12} \text{m}\).

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Key Concepts

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

Resolving Power
Resolving power is a key concept to understanding how well a diffraction grating can distinguish between closely spaced wavelengths. It tells us how capable the grating is at separating different wavelengths into distinct, observable lines. The formula to determine resolving power is \( R = nN \), where \( n \) represents the order of the spectrum, and \( N \) is the total number of lines on the grating.
  • When you have a higher number of lines \( N \), the resolving power increases. This means the grating is capable of resolving small differences in wavelength better.
  • The order of the spectrum \( n \) also plays a crucial role. For example, in higher orders like the second order, the resolving power will be higher.
In our case, we calculated a resolving power of 120,000 in the second-order spectrum. This high number indicates the grating can resolve very fine wavelength differences, making it very effective for detailed spectral analysis.
Wavelength Difference
The wavelength difference \( \Delta \lambda \) that a diffraction grating can resolve depends directly on its resolving power \( R \). It's the smallest difference in the wavelength that can be distinguished. This is given by the formula \( R = \frac{\lambda}{\Delta \lambda} \). To find the smallest wavelength difference \( \Delta \lambda \), simply rearrange this formula to get \( \Delta \lambda = \frac{\lambda}{R} \).
  • Here, \( \lambda \) is the mean wavelength, which in this situation is 540 nm (equivalent to 540 x 10-9 m).
  • A high resolving power means a small \( \Delta \lambda \).
For the scenario in question, given a resolving power of 120,000, the smallest resolvable wavelength difference is approximately 4.5 x 10-12 m. This implies that the grating can distinguish between wavelengths that are just fractions of a nanometer apart.
Second-Order Spectrum
The term "second-order spectrum" refers to the second set of diffraction patterns the light forms when it passes through a diffraction grating. For each wavelength entering the grating, there are several possible angles of diffraction dependent on the order of spectrum \( n \).
  • In higher orders like the second order (\( n = 2 \)), the light has been diffracted at sharper angles, which allows us to see more detailed and separated lines on the spectrum.
  • Higher order spectra, while providing more detailed data, can also lead to overlapping of different wavelengths if the grating resolving power is not sufficient.
In our specific example, utilizing the second-order spectrum considerably aids in achieving a high resolving power \, enabling more detailed analysis of the light's components. It highlights the finer details of dispersion, thus proving essential in applications like spectroscopy, where identifying specific wavelengths is critical.

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Most popular questions from this chapter

A beam of collimated polychromatic light ranging from \(500 \mathrm{nm}\) to \(700 \mathrm{nm}\) impinges normally on a transmission grating having 590000 lines/m. If the complete second-order spectrum is to appear, how wide, at most, can the slits be? [Hint: The second-order spectrum must fit within the diffraction envelope of each slit.

Imagine two aperture screens arranged to produce two Fraunhofer diffraction patterns. One contains 8 very narrow closely spaced parallel slits, the other 16 such slits. All else being equal, compare the two irradiance distributions. That is, how many subsidiary maxima between consecutive principal maxima will each pattern contain? If the irradiance of the zeroth-order peak of the 16 -slit pattern is set equal to \(1.0,\) how big will the corresponding peak be for the 8 -slit pattern? Which arrangement produces wider principal maxima? Draw a rough sketch of each.

Suppose that a grating spectrometer while in vacuum on Earth sends 500 -nm light off at an angle of \(20.0^{\circ}\) in the first-order spectrum. By comparison, after landing on the planet Mongo, the same light is diffracted through \(18.0^{\circ} .\) Determine the index of refraction of the Mongoian atmosphere.

A narrow single slit (in air) in an opaque screen is illuminated by infrared from a He-Ne laser at \(1152.2 \mathrm{nm}\), and it is found that the center of the tenth dark band in the Fraunhofer pattern lies at an angle of \(6.2^{\circ}\) off the central axis. Determine the width of the slit. At what angle will the tenth minimum appear if the entire arrangement is immersed in water \(\left(n_{w}=1.33\right)\) rather than air \(\left(n_{a}=1.00029\right) ?\)

Using Rayleigh's criterion, determine the smallest angle subtended by two points of equal brightness that can just be resolved by the human eye. Assume a pupil diameter of \(2.0 \mathrm{mm}\) and a mean wavelength of \(550 \mathrm{nm}\). The index of refraction of the medium within the eye is 1.337
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