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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 m from the slit. If the width of the central maximum is 6.00 mm, what is the slit width \(a\) if the wavelength is (a) 500 nm (visible light); (b) 50.0 \(\mu\)m (infrared radiation); (c) 0.500 nm (x rays)?

Short Answer

Expert verified
(a) 1.67 mm, (b) 4.17 mm, (c) 0.167 mm for slit width.

Step by step solution

01

Understanding Diffraction Pattern

The width of the central maximum in a single-slit diffraction pattern can be related to the slit width \(a\), the wavelength \(\lambda\), and the distance \(L\) from the slit to the screen by the formula for the angular width of the central maximum: \(\theta = \frac{\lambda}{a}\). For a small angle approximation, the width on the screen, \(w\), is given by \( w = 2L\tan\theta \approx 2L\theta \). Thus, \( a = \frac{2L\lambda}{w} \).
02

Convert Units Appropriately

We need all lengths in consistent units. Therefore, convert the given values of \(\lambda\) for each part: (a) 500 nm = 500 x 10^{-9} m, (b) 50.0 \(\mu\)m = 50.0 x 10^{-6} m, (c) 0.500 nm = 0.500 x 10^{-9} m. The width \(w = 6.00\) mm = 6.00 x 10^{-3} m, and \(L = 2.50\) m.
03

Calculate Slit Width for (a) Visible Light

Using the formula \( a = \frac{2L\lambda}{w} \) with \(\lambda = 500\times10^{-9}\ \text{m}\), \(L = 2.50\ \text{m}\), \(w = 6.00\times10^{-3}\ \text{m}\):\[a = \frac{2 \times 2.50 \times 500 \times 10^{-9}}{6.00 \times 10^{-3}} = \frac{2500 \times 10^{-9}}{3 \times 10^{-3}} = \frac{5.00 \times 10^{-6}}{3} = 1.67 \times 10^{-3} \ \text{m}\]
04

Calculate Slit Width for (b) Infrared Radiation

Using the same formula with \(\lambda = 50.0 \times 10^{-6}\ \text{m}\):\[a = \frac{2 \times 2.50 \times 50.0 \times 10^{-6}}{6.00 \times 10^{-3}} = \frac{250 \times 10^{-5}}{3 \times 10^{-3}} = \frac{12.5 \times 10^{-3}}{3} = 4.17 \times 10^{-3} \ \text{m}\]
05

Calculate Slit Width for (c) X-Rays

Using the same formula with \(\lambda = 0.500 \times 10^{-9}\ \text{m}\):\[a = \frac{2 \times 2.50 \times 0.500 \times 10^{-9}}{6.00 \times 10^{-3}} = \frac{2.50 \times 10^{-9}}{3 \times 10^{-3}} = \frac{0.500 \times 10^{-6}}{3} = 0.167 \times 10^{-3} \ \text{m}\]

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Key Concepts

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

Electromagnetic Radiation
Electromagnetic radiation refers to waves of electric and magnetic energy moving through space. It comes in various forms, depending on the frequency and wavelength. These forms include radio waves, microwaves, infrared radiation, visible light, ultraviolet light, X-rays, and gamma rays.
The type of electromagnetic radiation used in many physics experiments is monochromatic, meaning it has a single wavelength. For instance, visible light with a wavelength of 500 nm or infrared light at 50.0 碌m are examples. Each type of radiation has unique properties due to its wavelength.
Different wavelengths interact differently with materials and affect how these waves are observed in experiments, like the one described above where a beam passes through a narrow slit to create a diffraction pattern. Understanding these interactions is crucial in fields like optics and photonics.
Wavelength
Wavelength is the distance between two consecutive peaks of a wave. It's a vital property of waves, determining their type and how they interact with matter. In electromagnetic radiation, wavelength defines the energy and type of the wave.
Shorter wavelengths, such as X-rays ( frac{1}{2} nm), have higher energy, while longer wavelengths like infrared (50.0 碌m) are lower in energy. In optical physics, knowing the wavelength is crucial for understanding how light behaves when it meets obstacles, such as slits or lenses.
In the context of diffraction, wavelength helps determine the diffraction pattern, including the width of the central maximum observed when electromagnetic waves pass through a slit.
Single-slit Diffraction
In single-slit diffraction, waves spread after passing through a narrow opening. The light waves interfere, creating a pattern of dark and light bands. This is known as a diffraction pattern.
The central band, known as the central maximum, is the brightest and widest part. The size and shape of this pattern depend on the wavelength of the light and the width of the slit.
  • A wider slit results in a narrower central maximum.
  • A longer wavelength leads to a wider central maximum.
These characteristics can be predicted with mathematical formulas that relate the distance from the slit to the screen, the width of the slit, and the wavelength of the light.
Central Maximum
The central maximum is the most prominent feature in a diffraction pattern, occurring directly in line with the incoming waves. It's called a 'maximum' because it鈥檚 the brightest point in the diffraction pattern.
Its width is influenced by:
  • The wavelength of the light: Longer wavelengths create wider central maxima.
  • The width of the slit: Narrower slits result in wider central maxima.
  • The distance to the screen: Greater distances amplify the effect on observed width.
In practical applications, like those seen in the problem, knowing the width of the central maximum allows for the calculation of the slit width when other properties are known. Understanding these relationships is key to mastering concepts in wave optics.

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Most popular questions from this chapter

If you can read the bottom row of your doctor's eye chart, your eye has a resolving power of 1 arcminute, equal to \(1\over{60}\) degree. If this resolving power is diffraction limited, to what effective diameter of your eye's optical system does this correspond? Use Rayleigh's criterion and assume \(\lambda\) = 550 nm.

The wavelength range of the visible spectrum is approximately 380-750 nm. White light falls at normal incidence on a diffraction grating that has 350 slits/mm. Find the angular width of the visible spectrum in (a) the first order and (b) the third order. (\(Note\): An advantage of working in higher orders is the greater angular spread and better resolution. A disadvantage is the overlapping of different orders, as shown in Example 36.4.)

When the light is passed through the bottom of the sample container, the interference maximum is observed to be at 41\(^\circ\); when it is passed through the top, the corresponding maximum is at 37\(^\circ\). What is the best explanation for this observation? (a) The microspheres are more tightly packed at the bottom, because they tend to settle in the suspension. (b) The microspheres aremore tightly packed at the top, because they tend to float to the top of the suspension. (c) The increased pressure at the bottom makes the microspheres smaller there. (d) The maximum at the bottom corresponds to \(m\) = 2, whereas the maximum at the top corresponds to \(m\) = 1.

Your physics study partner tells you that the width of the central bright band in a single-slit diffraction pattern is inversely proportional to the width of the slit. This means that the width of the central maximum increases when the width of the slit decreases. The claim seems counterintuitive to you, so you make measurements to test it. You shine monochromatic laser light with wavelength \(\lambda\) onto a very narrow slit of width \(a\) and measure the width \(w\) of the central maximum in the diffraction pattern that is produced on a screen 1.50 m from the slit. (By "width," you mean the distance on the screen between the two minima on either side of the central maximum.) Your measurements are given in the table. (a) If \(w\) is inversely proportional to \(a\), then the product \(aw\) is constant, independent of \(a\). For the data in the table, graph \(aw\) versus \(a\). Explain why \(aw\) is not constant for smaller values of \(a\). (b) Use your graph in part (a) to calculate the wavelength \(\lambda\) of the laser light. (c) What is the angular position of the first minimum in the diffraction pattern for (i) \(a\) = 0.78 \(\mu\)m and (ii) \(a\) = 15.60 \(\mu\)m?

Light of wavelength 585 nm falls on a slit 0.0666 mm wide. (a) On a very large and distant screen, how many \(totally\) dark fringes (indicating complete cancellation) will there be, including both sides of the central bright spot? Solve this problem \(without\) calculating all the angles! (\(Hint\): What is the largest that sin \theta can be? What does this tell you is the largest that \(m\) can be?) (b) At what angle will the dark fringe that is most distant from the central bright fringe occur?
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