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Chapter 36: Problem 8

Monochromatic electromagnetic radiation with wavelength \(\lambda\) from a distant source passes through a slit. The diffraction pattern is observed on a screen 2.50 \(\mathrm{m}\) from the slit. If the width of the central maximum is 6.00 \(\mathrm{mm}\) , what is the slit width \(a\) if the wavelength is (a) 500 \(\mathrm{nm}\) (visible light); (b) 50.0\(\mu \mathrm{m}\) (infrared radiation); (c) 0.500 \(\mathrm{nm}(\mathrm{x}\) rays)?

Short Answer

Expert verified
(a) 0.417 mm, (b) 4.17 mm, (c) 0.0417 mm.

Step by step solution

01

Understand the Diffraction Formula

In a single-slit diffraction pattern, the angle to the first minimum on either side of the central maximum, \( \theta \), is given by \( a \sin \theta = m \lambda \), where \( a \) is the slit width, \( m \) is the order of the minimum (\( m = 1 \) for the first minimum), and \( \lambda \) is the wavelength of the light. The central maximum width \( W \) is related to the angle by the trigonometric approximation: \( W = 2L \tan \theta \), where \( L \) is the distance to the screen.
02

Relate Width of Central Maximum to Slit and Wavelength

The central maximum width is twice the distance from the center to the first minimum. Thus, \( W = 2L \tan \theta \approx 2L \sin \theta \) for small angles, which gives \( \sin \theta = \frac{W}{2L} \). Substitute this back into the diffraction equation: \( a \frac{W}{2L} = \lambda \).
03

Solve for Slit Width (a) for Different Wavelengths

Rearrange the formula to solve for the slit width: \( a = \frac{2L\lambda}{W} \). Use given values of \( \lambda \) for each part:(a) For \( \lambda = 500 \) nm, or \( 500 \times 10^{-9} \) m: \( a = \frac{2 \times 2.5 \times 500 \times 10^{-9}}{6.00 \times 10^{-3}} = 4.17 \times 10^{-4} \) m or 0.417 mm.(b) For \( \lambda = 50.0 \mu \text{m} \), or \( 50.0 \times 10^{-6} \) m: \( a = \frac{2 \times 2.5 \times 50.0 \times 10^{-6}}{6.00 \times 10^{-3}} = 4.17 \times 10^{-3} \) m or 4.17 mm.(c) For \( \lambda = 0.500 \) nm, or \( 0.500 \times 10^{-9} \) m: \( a = \frac{2 \times 2.5 \times 0.500 \times 10^{-9}}{6.00 \times 10^{-3}} = 4.17 \times 10^{-5} \) m or 0.0417 mm.

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Key Concepts

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

Central Maximum
In single-slit diffraction, the central maximum refers to the brightest and widest part of the diffraction pattern seen on the screen. It is important because it is the primary region where the diffracted light concentrates.
This area of high intensity comes from the constructive interference of light waves passing through the slit.
It is flanked by less intense bands known as secondary maxima and dark bands called minima.
Unlike the secondary maxima, this central maximum is much wider and brighter.
  • The width of the central maximum is directly related to how light bends, a result of diffraction.
  • Its width increases as the slit width decreases or as the wavelength of light increases.
Small angles are typically involved in central maximum calculations, allowing use of approximations like \( \tan \theta \approx \sin \theta \).
Understanding the central maximum illuminates why different colors or wavelengths bend different amounts and why certain diffraction patterns are seen.
Wavelength
The wavelength of light is a key factor in understanding diffraction patterns.
It is the distance between two consecutive peaks (or troughs) of a wave.
Light with a longer wavelength, like infrared, bends more around obstacles than light with a shorter wavelength, such as visible light.
  • Bigger wavelengths lead to a wider central maximum.
  • As seen in the problem, varying the wavelength from visible to infrared or x-rays changes the slit width needed for the same maximum width.
Wavelengths are usually measured in nanometers (nm) or micrometers (碌m).
The varying behavior of light in diffraction patterns showcases its wave-like properties, dictated by its wavelength.
Slit Width
The slit width \(a\) is the physical opening through which light passes and is crucial to diffraction patterns.
In single-slit diffraction, it determines how much and how light will spread out upon passing through.
The slit acts like a new source of waves; hence, the characteristics of these waves are affected by the slit width.
  • A narrower slit results in a wider central maximum because the light spreads out more.
  • It is inversely proportional to the central maximum width, so reducing \(a\) increases the pattern width.
The slit width is related to wavelength in the formula \(a = \frac{2L\lambda}{W}\), where \(L\) is the distance to the screen, \(\lambda\) is the wavelength, and \(W\) is the central maximum width.
Manipulating the slit width allows control over the diffraction pattern, illustrating the interplay between physical dimensions and wave properties.

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Most popular questions from this chapter

Resolution of the Eye. The maximum resolution of the eye depends on the diameter of the opening of the pupil (a diffraction effect) and the size of the retinal cells. The size of the retinal cells (about 5.0\(\mu \mathrm{m}\) in diameter) limits the size of an object at the near point \((25 \mathrm{cm})\) of the eye to a height of about 50\(\mu \mathrm{m} .\) To get a reasonable estimate without having to go through complicated calculations, we shall ignore the effect of the fluid in the eye.) (a) Given that the diameter of the human pupil is about 2.0 mm, does the Rayleigh criterion allow us to resolve a 50 -\mum-tall object at 25 \(\mathrm{cm}\) from the eye with light of wavelength 550 \(\mathrm{nm}\) ? (b) According to the Rayleigh criterion, what is the shortest object we could resolve at the \(25-\mathrm{cm}\) near point with light of wavelength 550 \(\mathrm{nm} ?(\mathrm{c})\) What angle would the object in part (b) subtend at the eye? Express your answer in minutes \(\left(60 \min =1^{\circ}\right),\) and compare it with the experimental value of about 1 min. (d) Which effect is more important in limiting the resolution of our eyes: diffraction or the size of the retinal cells?

Monochromatic light of wavelength 580 nm passes through a single slit and the diffraction pattern is observed on a screen. Both the source and screen are far enough from the slit for Fraunhofer diffraction to apply.(a) If the first diffraction minima are at \(\pm 90.0^{\circ},\) so the central maximum completely fills the screen, what is the width of the slit? (b) For the width of the slit as calculated in part (a), what is the ratio of the intensity at \(\theta=45.0^{\circ}\) to the intensity at \(\theta=0 ?\)

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?

A slit 0.360 \(\mathrm{mm}\) wide is illuminated by parallel rays of light that have a wavelength of 540 \(\mathrm{nm}\) . The diffraction pattern is observed on a screen that is 1.20 \(\mathrm{m}\) from the slit. The intensity at the center of the central maximum \(\left(\theta=0^{\circ}\right)\) is \(I_{0}\) . (a) What is the distance on the screen from the center of the central maximum to the first minimum? (b) What is the distance on the screen from the center of the central maximum to the point where the intensity has fallen to \(I_{0} / 2 ?\)

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 save- length \(\lambda\) . 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 \(+y\) -direction, perpendicular to the line of the antennas. Show that if \(d<\lambda,\) this is the only principal interference maximum 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 $$\theta=\arcsin \left(\frac{\delta \lambda}{2 \pi d}\right)$$ 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 beam can be made to sweep back and forth around \(\theta=0 .(\mathrm{c})\) 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 airplane's direction of flight? Give your answer in radians.
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