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Chapter 36: Problem 4

Light of wavelength 633 \(\mathrm{nm}\) from a distant source is incident on a slit 0.750 \(\mathrm{mm}\) wide, and the resulting diffraction pattern is observed on a screen 3.50 \(\mathrm{m}\) away. What is the distance between the two dark fringes on either side of the central bright fringe?

Short Answer

Expert verified
The distance is 5.908 mm.

Step by step solution

01

Identify the Diffraction Formula

For a single slit diffraction pattern, the position of the dark fringes is given by the formula: \( a \sin \theta = m\lambda \), where \( a \) is the slit width, \( \theta \) is the angle from the central maximum to the dark fringe, \( m \) is the order of the dark fringe, and \( \lambda \) is the wavelength of the light.
02

Apply the Small Angle Approximation

For small angles, \( \sin \theta \approx \theta \approx \tan \theta = \frac{y}{L} \), where \( y \) is the fringe distance from the central maximum on the screen, and \( L \) is the distance from the slit to the screen. Thus, \( a \frac{y}{L} = m\lambda \) simplifies to \( y = \frac{m\lambda L}{a} \).
03

Calculate the Position of the First-Order Dark Fringe on One Side

Substitute the given values: \( a = 0.750 \, \text{mm} = 0.750 \times 10^{-3} \, \text{m} \), \( \lambda = 633 \, \text{nm} = 633 \times 10^{-9} \, \text{m} \), \( L = 3.50 \, \text{m} \), and \( m = 1 \) for the first dark fringe. \( y_1 = \frac{(1)(633 \times 10^{-9})(3.50)}{0.750 \times 10^{-3}} \).
04

Compute the Distance Calculation

Perform the calculation to find \( y_1 \):\[y_1 = \frac{633 \times 10^{-9} \times 3.50}{0.750 \times 10^{-3}} = 2.954 \times 10^{-3} \, \text{m} = 2.954 \, \text{mm}\].
05

Double the Distance to Obtain Total Dark Fringe Distance

Since the problem asks for the total distance between two dark fringes on either side of the central bright fringe, double the calculated value: \( 2 \times y_1 = 2 \times 2.954 \, \text{mm} = 5.908 \, \text{mm} \).

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Key Concepts

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

Wavelength of Light
Light, as it travels, can be described by its wavelength, which is the distance between consecutive peaks of a wave. The wavelength of light determines its color. In this particular exercise, the light has a wavelength of 633 nanometers (nm).
  • Wavelength is a crucial factor in diffraction patterns, as it affects how light interacts with obstacles like slits.
  • Visible light falls within a range of about 400 nm to 700 nm, with 633 nm appearing as a red light to the human eye.
When light passes through a single slit, like in this exercise, the wavelength is used to determine the positions of bright and dark fringes on a screen.
This is essential in calculations involving diffraction patterns.
Single Slit Diffraction
Single slit diffraction occurs when light passes through a narrow opening and bends, creating a pattern of bright and dark areas on a screen. The slit acts as a new source of waves.
  • The diffraction pattern consists of a central bright spot, known as the central maximum, with alternating dark and bright fringes on either side.
  • The diffraction formula for single slits, \( a \sin \theta = m\lambda \), helps predict the angles where dark fringes will appear.
Here, \( a \) is the slit width, \( \theta \) is the angle relative to the screen, \( m \) is the order of the dark fringe, and \( \lambda \) is the light's wavelength.
This formula is foundational for predicting how light will spread and interfere in such setups.
Dark and Bright Fringes
In a diffraction pattern, the dark and bright fringes are the result of interference between waves.
  • Bright fringes are areas where waves constructively interfere, meaning their peaks align to amplify light.
  • Dark fringes are produced by destructive interference, where peaks and troughs offset each other, reducing light intensity.
The central bright fringe is the most intense, located directly in line with the light source and slit. The order of dark fringe (\( m \)) corresponds to the sequence of dark bands moving outward from the central bright fringe.
Understanding the positioning of these fringes allows us to calculate distances, like the total distance between two dark fringes on either side of the central fringe.
Small Angle Approximation
The small angle approximation simplifies calculations in optics and other wave phenomena. It is particularly useful when dealing with angles that are small.
For small angles, it is assumed \( \sin \theta \approx \theta \approx \tan \theta \), making mathematical evaluations easier.
  • This approximation transforms the equation for the position of dark fringes into \( y = \frac{m\lambda L}{a} \), where \( y \) is the fringe distance, \( L \) is the screen distance, and \( a \) and \( \lambda \) remain the slit dimension and wavelength respectively.
  • Utilizing this simplification allows for straightforward computations while maintaining the accuracy needed for practical problem solving.
In this case, it allows us to calculate the distance between dark fringes without complex calculations.

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Most popular questions from this chapter

Monochromatic light with wavelength 620 \(\mathrm{nm}\) passes through a circular aperture with diameter 7.4\(\mu \mathrm{m}\) . The resulting diffraction pattern is observed on a screen that is 4.5 \(\mathrm{m}\) from the aperture. What is the diameter of the Airy disk on the screen?

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 individual slit, \(\phi=(2 \pi d \sin \theta) / \lambda, \theta\) is the angle of the rays reaching \(P(\text { as measured from the perpendicular bisector of the slit arrangement }\)), 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 $$ e^{i k}=\cos z+i \sin z $$ where \(i=\sqrt{-1}\) . In this expression, \(\cos z\) is the real part of the complex number \(e^{i k},\) 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 $$ \sum_{n=0}^{N-1} E_{0} e^{i(k R-\omega+n \phi)} $$ (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 $$ E_{0}\left(\frac{e^{i N \phi}-1}{e^{i \phi}-1}\right) e^{i(k R-\omega t)}=E_{0}\left(\frac{e^{i N \phi / 2}-e^{-i N \phi / 2}}{e^{i \phi / 2}-e^{-i \phi / 2}}\right) e^{i(k R-\omega t+(N-1) \phi / 2]} $$ Then, using the relationship \(e^{i k}=\cos z+i \sin z,\) show that the (real) electric field at point \(P\) is $$ E_{p}(t)=\left[E_{0} \frac{\sin (N \phi / 2)}{\sin (\phi / 2)}\right] \cos [k R-\omega t+(N-1) \phi / 2] $$ The quantity in the first square brackets in this expression is the amphitude of the electric field at \(P .\) (d) Use the result for the electric-field amplitude in part (c) to show that the intensity at an angle \(\theta\) 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 interference, by a factor of \(4 .\) (Hint: Is \(I_{0}\) defined in the same way in both expressions?)

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?

Monochromatic light of wavelength \(\lambda=620 \mathrm{nm}\) from a distant source passes through a slit 0.450 \(\mathrm{mm}\) wide. The diffraction pattern is observed on a screen 3.00 \(\mathrm{m}\) from the slit. In terms of the intensity \(I_{0}\) at the peak of the central maximum, what is the intensity of the light at the screen the following distances from the center of the central maximum: (a) \(1.00 \mathrm{mm} ;\) (b) 3.00 \(\mathrm{mm}\) ; (c) 5.00 \(\mathrm{mm}\) ?

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.
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