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The \(D\) line in the spectrum of sodium is a doublet with wavelengths \(589.0\) and \(589.6 \mathrm{~nm} .\) Calculate the minimum number of lines needed in a grating that will resolve this doublet in the second-order spectrum.

Short Answer

Expert verified
A minimum of 492 lines is needed in the grating.

Step by step solution

01

Understand the Problem

We need to determine the minimum number of lines in a diffraction grating that can resolve two close wavelengths in the sodium spectrum, specifically a doublet with wavelengths 589.0 nm and 589.6 nm, in the second-order spectrum.
02

Review the Resolving Power Formula

The resolving power (R) of a grating for a given wavelength is given by \(R = \frac{\lambda}{\Delta \lambda} = mN\), where \(m\) is the order of the spectrum, \(N\) is the number of lines in the grating, \(\lambda\) is the average wavelength, and \(\Delta \lambda\) is the difference in wavelengths to be resolved.
03

Calculate Wavelength Difference

Determine \(\Delta \lambda\), the difference in wavelengths: \(\Delta \lambda = 589.6 - 589.0 = 0.6\, \text{nm}\).
04

Calculate Average Wavelength

Compute the average wavelength \(\lambda\) of the doublet: \(\lambda = \frac{589.0 + 589.6}{2} = 589.3\, \text{nm}\).
05

Apply Formula to Find Minimum N

Substitute \(m = 2\), \(\lambda = 589.3\, \text{nm}\), and \(\Delta \lambda = 0.6\, \text{nm}\) into the resolving power formula: \(R = \frac{589.3}{0.6} = 2N\). Solving for \(N\), we have \(N = \frac{589.3}{1.2}\).
06

Solve for N

Calculate the value of \(N\): \(N = \frac{589.3}{1.2} \approx 491.1\). Since \(N\) must be an integer, round up to the nearest whole number, giving \(N = 492\).

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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 when it comes to diffraction gratings and their ability to distinguish two close wavelengths. Think of it as a grating's capability to separate the colors of light into distinct lines. The resolving power, denoted as \( R \), can be calculated using the formula \( R = \frac{\lambda}{\Delta \lambda} = mN \). Here, \( \lambda \) is the average wavelength, \( \Delta \lambda \) is the wavelength difference, \( m \) represents the diffraction order, and \( N \) is the number of lines on the grating.

  • Higher resolving power means better separation of closely spaced spectral lines.
  • In practical scenarios, a higher number of lines (\( N \)) results in a better capability of the grating to resolve fine details in the spectrum.
Understanding resolving power is essential because it dictates how effective a diffraction grating will be for a particular application. It's about optimizing the grating to make sure it can clearly distinguish between very close wavelengths.
Wavelength Difference
Wavelength difference, or \( \Delta \lambda \), is simply the gap between two wavelengths in the spectrum that you wish to differentiate. It's a crucial part of assessing how challenging it might be for a grating to resolve those wavelengths.

When looking at the sodium doublet for instance, the wavelengths are \( 589.0 \text{ nm} \) and \( 589.6 \text{ nm} \). The difference, \( \Delta \lambda \), is calculated as \( 589.6 - 589.0 = 0.6 \text{ nm} \).

  • The smaller the \( \Delta \lambda \), the more difficult it is to resolve the wavelengths, requiring a higher resolving power from the grating.
  • This is why understanding and calculating \( \Delta \lambda \) is essential when designing or choosing a diffraction grating for experiments or devices.
In essence, understanding the wavelength difference allows you to gauge the difficulty of the task your grating will perform, forming a basis for choosing the correct size and specifications.
Second-order Spectrum
The second-order spectrum in diffraction refers to the scenario where the light not only bends to form outputs but does so at an angle that results in additional spectral lines, distinct from the primary or first order. It involves looking at the pattern of light in a sequence that extends multiple times the wavelength.

In the sodium doublet case, a second-order spectrum means that the order \( m \) is 2. This double diffraction increases the resolving power — effectively doubling it because \( m = 2 \) in the formula \( R = mN \).

  • Each increase in order leads to a potential increase in resolving power because more bending of light occurs.
  • However, the further from the central maximum, the weaker the intensity. Hence, there is a need for careful balancing.
The second-order spectrum gives scientists more opportunities to observe and resolve details that might be hidden in the primary spectrum, enhancing analysis precision in experiments and applications.

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Most popular questions from this chapter

Suppose that the central diffraction envelope of a double-slit diffraction pattern contains 11 bright fringes and the first diffraction minima eliminate (are coincident with) bright fringes. How many bright fringes lie between the first and second minima of the diffraction envelope?

Monochromatic light with wavelength \(538 \mathrm{~nm}\) is incident on a slit with width \(0.025 \mathrm{~mm}\). The distance from the slit to a screen is \(3.5\) \(\mathrm{m}\). Consider a point on the screen \(1.1 \mathrm{~cm}\) from the central maximum. Calculate (a) \(\theta\) for that point, (b) \(\alpha\), and (c) the ratio of the intensity at that point to the intensity at the central maximum.

A grating has 400 lines/mm. How many orders of the entire visible spectrum \((400-700 \mathrm{~nm})\) can it produce in a diffraction experiment, in addition to the \(m=0\) order?

In an experiment to monitor the Moon's surface with a light beam, pulsed radiation from a ruby laser \((\lambda=0.69 \mu \mathrm{m})\) was directed to the Moon through a reflecting telescope with a mirror radius of \(1.3 \mathrm{~m}\). A reflector on the Moon behaved like a circular flat mirror with radius \(10 \mathrm{~cm}\), reflecting the light directly back toward the telescope on Earth. The reflected light was then detected after being brought to a focus by this telescope. Approximately what fraction of the original light energy was picked up by the detector? Assume that for each direction of travel all the energy is in the central diffraction peak.

Suppose that two points are separated by \(2.0 \mathrm{~cm}\). If they are viewed by an eye with a pupil opening of \(5.0 \mathrm{~mm}\), what distance from the viewer puts them at the Rayleigh limit of resolution? Assume a light wavelength of \(500 \mathrm{~nm}\).

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