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Visible light passes through a diffraction grating that has 900 slits \(/ \mathrm{cm},\) and the interference pattern is observed on a screen that is 2.50 \(\mathrm{m}\) from the grating. (a) Is the angular position of the first- order spectrum small enough for \(\sin \theta \approx \theta\) to be a good approximation? (b) In the first-order spectrum, the maxima for two different wavelengths are separated on the screen by 3.00 \(\mathrm{mm}\) . What is the difference in these wavelengths?

Short Answer

Expert verified
(a) No, the small angle approximation is not valid. (b) The wavelength difference is 1.33 nm.

Step by step solution

01

Determine slit separation

First, we need to find the separation between the slits in the diffraction grating. Given that there are 900 slits per cm, the separation (also called lattice constant) in meters is calculated as \[d = \frac{1}{900 \times 100}\approx 1.11 \times 10^{-6} \ m\]
02

Use diffraction grating formula for first-order spectrum

The formula for the angles of maxima in a diffraction grating is given by \[d \cdot \sin(\theta) = m \cdot \lambda\] where \(d\) is the slit separation, \(\theta\) is the angle of diffraction, \(m\) is the order number, and \(\lambda\) is the wavelength of light. For first-order (\(m=1\)), we have \[1.11 \times 10^{-6} \cdot \sin(\theta) = \lambda.\]
03

Check if angular position is small

To check if \(\sin \theta \approx \theta\), we calculate\[ \theta \approx \frac{\lambda}{1.11 \times 10^{-6}} \] Assuming visible light has a mean wavelength of \(550 \ nm\), \[\theta \approx \frac{550 \times 10^{-9}}{1.11 \times 10^{-6}} \approx 0.495\ rad \approx 28.37^\circ\] Since \(\theta\) is not very small, the approximation \(\sin \theta \approx \theta\) may not hold well.
04

Relate distance, angle, and screen separation

The difference in the path between two wavelengths on the screen is given by \[\Delta y = L \cdot \Delta \theta \] where \(L = 2.5 \ m\) is the distance to the screen and \(\Delta y = 3.00 \times 10^{-3} \ m\). Thus,\[\Delta \theta = \frac{3.00 \times 10^{-3}}{2.5} = 1.2 \times 10^{-3} \ rad.\]
05

Calculate difference in wavelengths

Using small angle approximation \(\Delta \theta = \frac{\Delta \lambda}{d}\), \[\Delta \lambda = d \cdot \Delta \theta = 1.11 \times 10^{-6} \cdot 1.2 \times 10^{-3} = 1.33 \times 10^{-9} \ m \approx 1.33 \ nm.\]

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Key Concepts

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

first-order spectrum
When light passes through a diffraction grating, it creates an interference pattern that is composed of several bright spots known as spectra. The first-order spectrum refers to the first occurrence of a maximum on either side of the central maximum (zero-order maximum). It indicates that the path difference between light waves from consecutive slits is exactly one wavelength (\(m = 1\)). Light is diffracted at this angle where the condition \(d \cdot \sin(\theta) = \lambda\) is satisfied, with \(d\) being the slit separation and \(\lambda\) the wavelength of light. The first-order spectrum is especially important because it often provides the clearest and simplest measurement for analysis and calculations. Scientists use it to determine specific wavelengths of light, such as in spectroscopy, which can help identify substances based on their emission or absorption spectra.
wavelength separation
Wavelength separation is a key concept when analyzing the interference pattern produced by a diffraction grating. It quantifies the difference in wavelengths (\(\Delta \lambda\)) of two nearby spectral lines observed on a screen. In the context given, the separation between maxima on a screen, due to different wavelengths of light, can be represented as \(\Delta y\). The relationship is given by \(\Delta y = L \cdot \Delta \theta\), where \(L\) is the distance to the screen and \(\Delta \theta\) is the change in angle observed on the screen. This allows us to calculate the difference in the wavelengths through \(\Delta \lambda = d \cdot \Delta \theta\). By understanding wavelength separation, scientists can distinguish between different chemical elements or compounds emitting light at slightly different wavelengths. This is also crucial in optical engineering where separating light into its component wavelengths is fundamental.
diffraction grating formula
The diffraction grating formula is vital in determining how light interacts with a diffraction grating. The formula gives the condition for bright interference maxima and is expressed as \(d \cdot \sin(\theta) = m \cdot \lambda\). Here, \(d\) is the slit separation, \(\theta\) is the diffraction angle, \(m\) is the order of the maximum, and \(\lambda\) is the wavelength of light.This formula helps predict the angles at which light will constructively interfere, forming a bright band on a screen. Different orders (\(m\)) represent the different levels of the spectrum observed. By using this formula, one can determine the wavelength if the angle and other parameters are known, or find the angle if the wavelength is given. Understanding and applying this formula allows for precise control and measurement in various optical applications, such as in spectrometers used in labs to identify substances.
angular position approximation
In diffraction experiments, especially with gratings, it is sometimes helpful to approximate \(\sin(\theta)\) as \(\theta\) for very small angles. This is the angular small-angle approximation. It simplifies calculations significantly because trigonometric functions (like sine) become unnecessary.The approximation \(\sin \theta \approx \theta\) (in radians) holds when \(\theta\) is small, generally for angles less than about 10 degrees. This approximation allows simplification of the diffraction formula to \(\theta \approx \frac{\lambda}{d}\), provided the angle is indeed sufficiently small.In practice, this approximation aids in quickly calculating expected spectral positions without complex trigonometric evaluations, especially in educational settings. When dealing with larger angles, one must use the full trigonometric expression as the assumption no longer holds valid, to ensure accuracy.

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Most popular questions from this chapter

The intensity of light in the Fraunhofer diffraction pattern of a single slit is $$ I=I_{0}\left(\frac{\sin \gamma}{\gamma}\right)^{2} $$ where $$ \gamma=\frac{\pi a \sin \theta}{\lambda} $$ (a) Show that the equation for the values of \(\gamma\) at which \(I\) is a maximum is \(\tan \gamma=\gamma\) (b) Determine the three smallest positive values of \(\gamma\) that are solutions of this equation (Hint: You can use a trial-and-error procedure. Guess a value of \(\gamma\) and adjust your guess to bring tan \(\gamma\) closer to \(\gamma\) A graphical solution of the equation is very helpful in locating the solutions approximately, to get good initial guesses.)

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.

The light from an iron are 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?

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?

Laser light of wavelength 632.8 \(\mathrm{nm}\) falls normally on a slit that is 0.0250 \(\mathrm{mm}\) wide. The transmitted light is viewed on a distant screen where the intensity at the center of the central bright fringe is \(8.50 \mathrm{W} / \mathrm{m}^{2} .\) (a) Find the maximum number of totally dark fringes on the screen, assuming the screen is large enough to show them all. (b) At what angle does the dark fringe that is most distant from the center occur? (c) What is the maximum intensity of the bright fringe that occurs immediately before the dark fringe in part (b)? Approximate the angle at which this fringe occurs by assuming it is midway between the angles to the dark fringes on either side of it.

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