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A diffraction grating has 650 slits \(/ \mathrm{mm}\) . What is the highest order that contains the entire visible spectrum? (The wavelength range of the visible spectrum is approximately \(400-700 \mathrm{nm} . )\)

Short Answer

Expert verified
The highest order that contains the entire visible spectrum is 2.

Step by step solution

01

Calculate the Grating Spacing

The grating spacing, denoted as \( d \), is the distance between adjacent slits. It can be calculated by taking the reciprocal of the number of slits per millimeter. Thus, for a grating with 650 slits/mm, the spacing is \( d = \frac{1}{650} \ mm = \frac{10^{-3}}{650} \ m \approx 1.54 \times 10^{-6} \ m\).
02

Use the Diffraction Grating Formula

The formula for diffraction maxima is \( d \sin \theta = m \lambda \), where \( m \) is the order number, \( \lambda \) is the wavelength, and \( \theta \) is the angle of diffraction. To find the highest possible \( m \), we know that \( \sin \theta \leq 1 \). Thus, \( m = \frac{d}{\lambda} \).
03

Calculate Maximum Order for Visible Spectrum

We need to find \( m \) for the extreme wavelengths of the visible spectrum. For the longest wavelength, \( \lambda = 700 \ nm = 700 \times 10^{-9} \ m \). Using \( m = \frac{d}{\lambda} \), we get \( m_{max} = \frac{1.54 \times 10^{-6}}{700 \times 10^{-9}} \approx 2.2 \). Since \( m \) must be an integer, the maximum complete order is \( m = 2 \).

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Key Concepts

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

Grating Spacing
Grating spacing is a key concept when working with diffraction gratings. It refers to the distance between two adjacent slits in the grating. To calculate grating spacing, you take the reciprocal of the number of slits per unit length. For example, if a grating has 650 slits per millimeter, the spacing \( d \) is calculated as follows:
\[ d = \frac{1}{650} \ \, mm = \frac{10^{-3}}{650} \ \, m \approx 1.54 \times 10^{-6} \ \, m \]
This indicates that each slit is approximately \( 1.54 \times 10^{-6} \) meters apart.

Understanding grating spacing is crucial because it directly affects the diffraction pattern produced. The more slits per millimeter, the closer the spacing, allowing the grating to create more precise and detailed patterns. This distance plays a significant role in determining the angles at which different wavelengths of light will diffract, or spread out, into their component colors.
Visible Spectrum
The visible spectrum encompasses the range of electromagnetic wavelengths that are detectable by the human eye. This spectrum spans from approximately 400 nm to 700 nm.
These limits mark the shortest and longest wavelengths that we perceive as light, forming the colors from violet to red. Understanding the visible spectrum is essential in applications like spectroscopy, where different wavelengths need to be analyzed or measured.

The visible spectrum is a small part of the larger electromagnetic spectrum, which includes many other types of waves like ultraviolet or infrared. Each color within the visible range corresponds to a specific range of wavelengths. For instance:
  • Violet light appears around 400-450 nm.
  • Blue light ranges from 450-495 nm.
  • Green light appears from 495-570 nm.
  • Yellow light ranges from 570-590 nm.
  • Orange light appears from 590-620 nm.
  • Red light ranges from 620-700 nm.
Visualizing how these wavelengths are spread out helps us understand how a diffraction grating can separate white light into its constituent colors.
Diffraction Order
In the context of a diffraction grating, the term 'order' refers to the number of wavelengths that fit into the path difference between adjacent slits. This is represented by the integer \( m \) in the diffraction equation \( d \sin \theta = m \lambda \).
The diffraction order helps determine how many maxima, or peaks, can be produced for light of a given wavelength. It signifies the level of diffraction, indicating how many cycles of a single wavelength are spaced across the width of the grating.

The order affects how light is dispersed when it passes through the grating. Higher orders of diffraction (e.g., \( m = 2, 3 \)) will have wider spreading angles compared to the first order \( m = 1 \). However, only complete orders can display the entire visible spectrum without missing any colors, which is why finding the maximum order that accommodates every color in the visible spectrum is essential for such exercises.
Wavelength Range
The wavelength range in a diffraction setting refers to the array of wavelengths that can be affected or utilized within a particular experiment. For the visible spectrum, this range lies between 400 nm and 700 nm. This entire set of wavelengths must fit inside the diffraction orders available, for proper analysis.

When we analyze light with a diffraction grating, the range of visible wavelengths determines how many orders can fully display the light's spectrum. The aim is to find the highest diffraction order that can incorporate the entire range of these wavelengths. In the given problem:
  • The shortest wavelength is 400 nm.
  • The longest wavelength is 700 nm.
Using formulas and calculations, we can determine which orders cover this wavelength span. Understanding the wavelength range is crucial in estimating how much of the light's spectrum will be observed within a given order of diffraction.

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Most popular questions from this chapter

Phased-Array Radar. In one common type of radar installation, a rotating antenna sweeps a radio beam around the sky. But in a phased-array radar system, the antennas remain stationary and the beam is swept electronically. To see how this is done, consider an array of \(N\) antennas that are arranged along the horizontal \(x\) -axis at \(x=0, \pm d, \pm 2 d, \ldots, \pm(N-1) d / 2 .\) (The number \(N\) is odd.) Each antenna emits radiation uniformly in all directions in the horizontal \(x y\) -plane. The antennas all emit radiation coherently, with the same amplitude \(E_{0}\) and the same wavelength \(\lambda\) . The relative phase \(\delta\) of the emission from adjacent antennas can be varied, however. If the antenna at \(x=0\) emits a signal that is given by \(E_{0} \cos \omega t,\) as measured at a point next to the antenna, the antenna at \(x=d\) emits a signal given by \(E_{0} \cos (\omega t+\delta),\) as measured at a point next to that antenna. The corresponding quantity for the antenna at \(x=-d\) is \(E_{0} \cos (\omega t-\delta) ;\) for the antennas at \(x=\pm 2 d,\) it is \(E_{0} \cos (\omega t \pm 2 \delta) ;\) and so on. \((a)\) If \(\delta=0\) , the interference pattern at a distance from the antennas is large compared to \(d\) and has a principal maximum at \(\theta=0\) (that is, in the angular range \(-90^{\circ}<\theta<90^{\circ} .\) Hence this principal maximum describes a beam emitted in the direction \(\theta=0\) . As described in Section \(36.4,\) if \(N\) is large, the beam will have a large intensity and be quite narrow. (b) If \(\delta \neq 0\) , show that the principal intensity maximum described in part (a) is located at $$ \boldsymbol{\theta}=\arcsin \left(\frac{\delta \boldsymbol{\lambda}}{2 \pi d}\right) $$ where \(\delta\) is measured in radians. Thus, by varying \(\delta\) from positive to negative values and back again, which can easily be done electronically, the bearn can be made to sweep back and forth around \(\boldsymbol{\theta}=\mathbf{0} .\) (c) A weather radar unit to be installed on an airplane emits radio waves at 8800 \(\mathrm{MHz}\) . The unit uses 15 antennas in an array 28.0 \(\mathrm{cm}\) long (from the antenna at one end of the array to the antenna at the other end). What must the maximum and minimum values of \(\delta\) be (that is, the most positive and most negative values) if the radar beam is to sweep \(45^{\circ}\) to the left or right of the air- plane's direction of flight? Give your answer in radians.

Monochromatic light is at normal incidence on a plane transmission grating. The first-order maximum in the interference pattern is at an angle of \(8.94^{\circ} .\) What is the angular position of the fourth-order maximum?

An interference pattern is produced by light of wavelength 580 \(\mathrm{nm}\) from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.530 \(\mathrm{mm}\) . (a) If the slits are very narrow, what would be the angular positions of the first-order and second- order, two-slit, interference maxima? (b) Let the slits have width 0.320 \(\mathrm{mm}\) , In terms of the intensity \(I_{0}\) at the center of the central maximum, what is the intensity at each of the angular positions in part (a)?

If the planes of a crystal are 3.50 \(\mathrm{A}\left(\mathrm{i} \mathrm{A}=10^{-10} \mathrm{m}=\right.\) 1 Angstrom unit) apart, (a) what wavelength of electromagnetic waves is necded so that the first strong interference maximum in the Bragg reflection occurs when the waves strike the planes at an angle of \(15.0^{\circ}\) , and in what part of the electromagnetic spectrum do these waves lie? (See Fig. 32.4.) (b) At what other angles will strong interference maxima occur?

An interference pattern is produced by four parallel and equally spaced, narrow slits. By drawing appropriate phasor diagrams, show that there is an interference minimum when the phase difference \(\phi\) from adjacent slits is (a) \(\pi / 2 ;(b) \pi ;\) (c) 3\(\pi / 2\) . In each case, for which pairs of slits is there totally destructive interference?

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