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Chapter 36: Problem 38

The light from an iron arc includes many different wavelengths. Two of these are at \(\lambda=587.9782 \mathrm{nm}\) and \(\lambda=\) 587.8002 \(\mathrm{nm} .\) You wish to resolve these spectral lines in first order using a grating 1.20 \(\mathrm{cm}\) in length. What minimum number of slits per centimeter must the grating have?

Short Answer

Expert verified
The grating must have at least 2753 slits per centimeter.

Step by step solution

01

Understand the Problem

We are given two wavelengths of light from an iron arc: \( \lambda_1 = 587.9782 \, \text{nm} \) and \( \lambda_2 = 587.8002 \, \text{nm} \). We need to find the minimum number of slits per centimeter that a diffraction grating should have to just resolve these wavelengths in the first order.
02

Use Rayleigh's Criterion for Resolution

Rayleigh's criterion for resolving two wavelengths \( \lambda_1 \) and \( \lambda_2 \) using a diffraction grating is given by:\[ \Delta \lambda = \frac{\lambda}{N \cdot m} \]where \( \Delta \lambda = \lambda_1 - \lambda_2 \), \( N \) is the total number of lines on the grating, and \( m \) is the order of diffraction. Convert the wavelengths from nm to cm: \( \lambda_1 = 587.9782 \, \text{nm} = 5.879782 \, \text{x} \, 10^{-5} \, \text{cm} \) and \( \lambda_2 = 587.8002 \, \text{nm} = 5.878002 \, \text{x} \, 10^{-5} \, \text{cm} \). The difference \( \Delta \lambda \) in cm is \( 1.780 \, \text{x} \, 10^{-8} \, \text{cm} \).
03

Calculate Using Grating Formula

We need to resolve the lines in first order (\( m = 1 \)), thus:\[ N = \frac{\lambda}{\Delta \lambda \cdot m} = \frac{5.879782 \, \text{x} \, 10^{-5}}{1.780 \, \text{x} \, 10^{-8} \cdot 1} \approx 3303 \text{ lines} \].This is the total number of lines in the 1.20 cm length grating needed for resolution.
04

Calculate Slits per Centimeter

The grating is 1.20 cm long, so the number of slits per centimeter \( d \) is calculated as:\[ d = \frac{N}{1.20} \approx \frac{3303}{1.20} \approx 2752.5 \text{ slits/cm} \].We must round up to the nearest whole number to ensure resolution, so we require 2753 slits per centimeter as the minimum.

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Key Concepts

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

Rayleigh's criterion
Rayleigh's criterion is essential for understanding how we can resolve two close wavelengths of light using a diffraction grating. It states that two wavelengths will be just resolved if the principal maximum of one coincides with the first minimum of the other. This implies that the ability to distinguish two wavelengths depends on the separation of their diffraction fringes.
Rayleigh's criterion for a grating is given by the equation:
  • \( \Delta \lambda = \frac{\lambda}{N \cdot m} \)
This equation tells us the minimum wavelength difference \( \Delta \lambda \) that can be resolved, where \( \lambda \) is the average wavelength, \( N \) is the total number of lines in the grating, and \( m \) is the diffraction order.
Using Rayleigh's criterion, we can determine the number of lines required on a grating to just resolve two wavelengths of light. This is particularly useful in applications like spectroscopy, where precision in resolving different wavelengths is crucial.
spectral resolution
Spectral resolution refers to the ability of an optical instrument to separate close wavelengths. A higher spectral resolution implies the ability to distinctly separate two nearly identical wavelengths. The spectral resolution of a diffraction grating depends on several factors:
  • The number of slits or lines in the grating
  • The length of the grating
  • The order of diffraction being used
To achieve optimal spectral resolution, the grating must have a sufficient number of slits that produce sharp and distinct diffraction patterns. In the given problem, the resolution is determined by whether we can see both wavelengths clearly when they hit the grating. Calculating the necessary slits per centimeter ensures that the instrument has the proper resolving power.
As shown in the exercise, by determining the number of slits per centimeter required for resolving the specific wavelength difference, we can ascertain the spectral resolution. A calculation leading to 2753 slits per centimeter means that the grating can just resolve the given spectral lines.
slits per centimeter
Slits per centimeter on a diffraction grating are crucial for determining its resolving power. Greater slits per centimeter means a higher ability to differentiate between closely spaced spectral lines. In our exercise, the light grating鈥檚 property is calculated based on several components:
  • Length of the grating
  • Total number of lines
  • Desired order of resolution
To find the slits per centimeter (also known as line density), we divide the total number of lines \( N \) by the physical length of the grating in centimeters. Using the problem's data, we found that approximately 2753 slits per centimeter are needed. This number ensures that the grating can resolve the given wavelengths in the first order, according to Rayleigh's criterion.
In practical terms, having more slits per centimeter allows the grating to spread light over a larger range, increasing its resolving power. This is vital for applications like identifying fine structures in spectra, widely used in materials science and astronomy.

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Most popular questions from this chapter

A loudspeaker having a diaphragm that vibrates at 1250 \(\mathrm{Hz}\) is traveling at 80.0 \(\mathrm{m} / \mathrm{s}\) directly toward a pair of holes in a very large wall in a region for which the speed of sound is 344 \(\mathrm{m} / \mathrm{s} .\) You observe that the sound comind through the openings first cancels at \(\pm 11.4^{\circ}\) with respect to the original direction of the speaker when observed far from the wall. (a) How far apart are the two openings? (b) At what angles would the sound first cancel if the source stopped moving?

Intensity Pattern of \(N\) Slits. (a) Consider an arrangement of \(N\) slits with a distance \(d\) between adjacent slits. The slits emit coherently and in phase at wavelength \(\lambda .\) Show that at a time \(t,\) the electric field at a distant point \(P\) is $$\begin{aligned} E_{P}(t)=& E_{0} \cos (k R-\omega t)+E_{0} \cos (k R-\omega t+\phi) \\\ &+E_{0} \cos (k R-\omega t+2 \phi)+\cdots \\ &+E_{0} \cos (k R-\omega t+(N-1) \phi) \end{aligned}$$ where \(E_{0}\) is the amplitude at \(P\) of the electric field due to an indi- vidual slit, \(\phi=(2 \pi d \sin \theta) / \lambda, \theta\) is the angle of the rays reaching \(P\) (as measured from the perpendicular bisector of the slit arrange- ment), and \(R\) is the distance from \(P\) to the most distant slit. In this problem, assume that \(R\) is much larger than \(d\) . (b) To carry out the sum in part (a), it is convenient to use the complex-number relationship $$e^{i z}=\cos z+i \sin z$$ where \(i=\sqrt{-1}\) . In this expression, cos \(z\) is the real part of the complex number \(e^{i z},\) and \(\sin z\) is its imaginary part. Show that the electric field \(E_{P}(t)\) is equal to the real part of the complex quantity $$\sum_{n=0}^{N-1} E_{0} e^{i(k R-\omega t+n \phi)}$$ (c) Using the properties of the exponential function that \(e^{A} e^{B}=e^{(A+B)}\) and \(\left(e^{A}\right)^{n}=e^{n A},\) show that the sum in part (b) can be written as $$E_{0}\left(\frac{e^{i N \phi}-1}{e^{i \phi}-1}\right) e^{i(k R-\omega t)}\( \)\quad=E_{0}\left(\frac{e^{i N \phi / 2}-e^{-i N \phi / 2}}{e^{i \phi / 2}-e^{-i \phi / 2}}\right) e^{i[k R-\omega t+(N-1) \phi}$$ Then, using the relationship \(e^{i z}=\cos z+i \sin z,\) show that the (real) electric field at point \(P\) is $$I=I_{0}\left[\frac{\sin (N \phi / 2)}{\sin (\phi / 2)}\right]^{2}$$ where \(I_{0}\) is the maximum intensity for an individual slit. (e) Check the result in part (d) for the case \(N=2 .\) It will help to recall that \(\sin 2 A=2 \sin A \cos A .\) Explain why your result differs from Eq. \((35.10),\) the expression for the intensity in two-source interfer- ence, by a factor of \(4 .\) Hint: Is \(I_{0}\) defined in the same way in both expressions?

Parallel rays of monochromatic light with wavelength 568 nm illuminate two identical slits and produce an interference pattern on a screen that is 75.0 \(\mathrm{cm}\) from the slits. The centers of the slits are 0.640 \(\mathrm{mm}\) apart and the width of each slit is 0.434 \(\mathrm{mm}\) . If the intensity at the center of the central maximum is \(5.00 \times\) \(10^{-4} \mathrm{W} / \mathrm{m}^{2},\) what is the intensity at a point on the screen that is 0.900 \(\mathrm{mm}\) from the center of the central maximum?

Photography. A wildlife photographer uses a moderate telephoto lens of focal length 135 \(\mathrm{mm}\) and maximum aperture \(f / 4.00\) to photograph a bear that is 11.5 \(\mathrm{m}\) away. Assume the wavelength is 550 \(\mathrm{nm}\) (a) What is the width of the smallest feature on the bear that this lens can resolve if it is opened to its maximum aperture? (b) If, to gain depth of field, the photographer stops the lens down to \(f / 22.0,\) what would be the width of the smallest resolvable feature on the bear?

Measuring Refractive Index. A thin slit illuminated by light of frequency \(f\) produces its first dark band at \(\pm 38.2^{\circ}\) in air. When the entire apparatus (slit, screen, and space in between) is immersed in an unknown transparent liquid, the slit's first dark bands occur instead at \(\pm 21.6^{\circ} .\) Find the refractive index of the liquid.
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