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Chapter 36: Problem 60

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
The grating needs at least 492 lines to resolve the doublet.

Step by step solution

01

Formula for Resolving Power

The resolving power \( R \) of a diffraction grating is defined by the formula \( R = \dfrac{\lambda}{\Delta \lambda} = mN \), where \( \lambda \) is the average wavelength, \( \Delta \lambda \) is the difference in wavelength, \( m \) is the order of the spectrum, and \( N \) is the number of lines.
02

Calculate Average Wavelength

To find the average wavelength \( \lambda \), take the average of the two given wavelengths: \[ \lambda = \dfrac{589.0 + 589.6}{2} = 589.3 \ \text{nm}. \]
03

Calculate Wavelength Difference

The difference between the two wavelengths \( \Delta \lambda \) is calculated as: \[ \Delta \lambda = 589.6 - 589.0 = 0.6 \ \text{nm}. \]
04

Solving for Number of Lines

Substitute the values and the order \( m = 2 \) into the resolving power equation: \[ R = mN = \dfrac{\lambda}{\Delta \lambda} = \dfrac{589.3}{0.6} \approx 982.17. \] So, the number of lines needed, \( N \), is \[ N = \dfrac{982.17}{2} \approx 491.085. \]
05

Round to Nearest Whole Number

Since the number of lines \( N \) must be a whole number, we round 491.085 up to the nearest whole number, which gives: \[ N = 492. \]

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Key Concepts

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

Resolving Power in Diffraction Gratings
The resolving power of a diffraction grating is a crucial concept in optics. It helps determine how well a grating can distinguish between two nearby wavelengths.
  • Formula: The resolving power \( R \) is given by the equation \( R = \dfrac{\lambda}{\Delta \lambda} = mN \). This equation relates the average wavelength \( \lambda \), the wavelength difference \( \Delta \lambda \), the order of the spectrum \( m \), and the number of lines \( N \) on the grating.
  • Average Wavelength: To calculate \( R \), you first find the average of the two wavelengths, which is crucial for understanding the specific line you are resolving.
  • Importance: A higher resolving power means the grating can separate two wavelengths that are very close together, making it easier to identify different spectral lines.
In our sodium doublet example, the resolving power allows us to calculate how many lines are required on a grating to clearly separate the lines with wavelengths 589.0 nm and 589.6 nm in the second-order spectrum, which is \( m = 2 \).
Understanding Wavelength Doublet
A wavelength doublet consists of two closely spaced wavelengths, typically originating from the same element or molecule. Doublets occur due to slight differences in energy transitions of electrons.
  • Example: For sodium, the \( D \) line appearing as a doublet involves wavelengths of 589.0 nm and 589.6 nm. These lines result from the fine structure splitting of energy levels in sodium atoms.
  • Resolution: To distinguish doublets, optical instruments need to have sufficient resolving power to separate the two peaks into individual components.
  • Applications: Understanding and resolving doublets is vital in spectroscopic investigations, where accurate identification of substances depends on recognizing these closely lying spectral lines.
In practical usage, resolving such doublets, particularly in the field of astronomy and laboratory spectroscopy, helps identify different compounds and elements with precision.
Second-Order Spectrum
A second-order spectrum refers to the set of diffraction patterns generated when light waves constructively interfere at angles corresponding to the second multiple of the wavelength. This is denoted by \( m = 2 \).
  • Characteristic: The second-order spectrum typically appears at higher diffraction angles compared to the first-order. This can help improve the resolution of the spectrum.
  • Resolution Enhancement: Higher orders like the second-order can offer better resolution. This means it allows clearer separation of closely spaced wavelengths.
  • Practical Implication: Using the second-order spectrum can be beneficial in scenarios requiring enhanced resolution, such as in the sodium doublet example, where distinguishing between 589.0 nm and 589.6 nm is necessary.
The calculation in the exercise demonstrates how using the second-order of a diffraction grating involves a different line density to achieve the desired resolution. This distinguishes it from the first-order, emphasizing its importance in applications demanding finer spectral details.

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Most popular questions from this chapter

Assume that the limits of the visible spectrum are arbitrarily chosen as 430 and \(680 \mathrm{nm}\). Calculate the number of rulings per millimeter of a grating that will spread the first-order spectrum through an angle of \(20.0^{\circ}\).

The wall of a large room is covered with acoustic tile in which small holes are drilled \(5.0 \mathrm{~mm}\) from center to center. How far can a person be from such a tile and still distinguish the individual holes, assuming ideal conditions, the pupil diameter of the observer's eye to be \(4.0 \mathrm{~mm},\) and the wavelength of the room light to be \(550 \mathrm{nm} ?\)

A spy satellite orbiting at \(160 \mathrm{~km}\) above Earth's surface has a lens with a focal length of \(3.6 \mathrm{~m}\) and can resolve objects on the ground as small as \(30 \mathrm{~cm}\). For example, it can easily measure the size of an aircraft's air intake port. What is the effective diameter of the lens as determined by diffraction consideration alone? Assume \(\lambda=550 \mathrm{nm}\).

Light of wavelength \(600 \mathrm{nm}\) is incident normally on a diffraction grating. Two adjacent maxima occur at angles given by \(\sin \theta=0.2\) and \(\sin \theta=0.3 .\) The fourth-order maxima are missing. (a) What is the separation between adjacent slits? (b) What is the smallest slit width this grating can have? For that slit width, what are the (c) largest, (d) second largest, and (e) third largest values of the order number \(m\) of the maxima produced by the grating?

An astronaut in a space shuttle claims she can just barely resolve two point sources on Earth's surface, \(160 \mathrm{~km}\) below. Calculate their (a) angular and (b) linear separation, assuming ideal conditions. Take \(\lambda=540 \mathrm{nm}\) and the pupil diameter of the astronaut's eve to be \(5.0 \mathrm{~mm}\)
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