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Chapter 35: Problem 9

35.9. Two slits spaced 0.450 \(\mathrm{mm}\) apart are placed 75.0 \(\mathrm{cm}\) from a screen. What is the distance between the second and third dark lines of the interference pattern on the screen when the slits are illuminated with coherent light with a wavelength of 500 \(\mathrm{nm} ?\)

Short Answer

Expert verified
0.83 mm

Step by step solution

01

Identify Known Variables

First, identify the given values in the problem: the slit separation \( d = 0.450 \, \text{mm} = 0.450 \times 10^{-3} \, \text{m} \), the distance from the slits to the screen \( L = 75.0 \, \text{cm} = 0.750 \, \text{m} \), and the wavelength of the light \( \lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m} \). We need to find the distance between the second and third dark lines.
02

Determine Condition for Dark Fringes

In a double-slit experiment, dark fringes occur at positions where the path difference is \( (m + \frac{1}{2})\lambda \), where \( m \) is an integer (0, 1, 2,...). The position on the screen for the dark fringes is given by \( y_m = \left(m + \frac{1}{2}\right)\frac{\lambda L}{d} \).
03

Calculate Position of Second Dark Fringe

Use the formula for dark fringes to find the position of the second dark fringe (for \( m = 1 \)):\[ y_2 = \left(1 + \frac{1}{2}\right)\frac{\lambda L}{d} = \left(\frac{3}{2}\right)\frac{500 \times 10^{-9} \times 0.750}{0.450 \times 10^{-3}} \approx 1.25 \times 10^{-3} \, \text{m} = 1.25 \, \text{mm}. \]
04

Calculate Position of Third Dark Fringe

Now, calculate the position of the third dark fringe (for \( m = 2 \)):\[ y_3 = \left(2 + \frac{1}{2}\right)\frac{\lambda L}{d} = \left(\frac{5}{2}\right)\frac{500 \times 10^{-9} \times 0.750}{0.450 \times 10^{-3}} \approx 2.08 \times 10^{-3} \, \text{m} = 2.08 \, \text{mm}. \]
05

Determine Distance Between Second and Third Dark Lines

The distance between the second and third dark lines is the difference between their positions:\[ \Delta y = y_3 - y_2 = 2.08 \, \text{mm} - 1.25 \, \text{mm} = 0.83 \, \text{mm}. \]

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Key Concepts

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

Dark Fringes
In a double-slit interference pattern, dark fringes represent the locations where destructive interference occurs. At these spots, waves from the two slits meet out of phase, canceling each other out. This happens when the path difference between the two waves equals a half-integral number of wavelengths.
  • For dark fringes, the path difference must be \( (m + \frac{1}{2}) \lambda \), where \( m \) is an integer.
  • The position of the dark fringes on a screen is calculated by the formula: \( y_m = \left(m + \frac{1}{2}\right)\frac{\lambda L}{d} \).
  • Dark fringes are typically labeled as the order of the fringe, such as the first, second, third, etc.
In the given exercise, to find the specific position of dark fringes, you have to apply this formula and choose the correct integer value for \( m \). Understanding dark fringes helps you predict where dark spots will appear in such a pattern.
Path Difference
The path difference is a crucial concept in wave interference, especially in a double-slit experiment. It refers to the difference in the distance traveled by two waves from the slits to a specific point on the screen. This difference causes either constructive or destructive interference, creating bright or dark fringes respectively.
  • The path difference can be calculated as the difference between two distances from each slit to the point on the screen where interference occurs.
  • For dark fringes, the path difference must satisfy \( (m + \frac{1}{2}) \lambda \).
  • It directly affects the position of the fringe on the screen based on its value.
Mastering the path difference concept enables you to determine exactly where interference patterns will form. It's an essential factor in precisely locating the positions of all the fringes on the screen.
Wavelength
Wavelength, denoted by \( \lambda \), is the distance between consecutive peaks (or troughs) of a wave. It is a fundamental characteristic not only in light waves but across all types of waves, including sound and water waves. In the context of optical interference from double slits, wavelength plays a pivotal role.
  • The specific wavelength of light used dictates how closely together the interference fringes are spaced.
  • Shorter wavelengths result in more closely spaced fringes, while longer wavelengths spread them further apart.
  • In experiments, it's usually given in nanometers (nm), where 1 nm = \( 10^{-9} \) meters, as seen in the exercise where \( \lambda = 500 \ \text{nm} \).
Knowing the wavelength allows you to predict the pattern of interference and calculate positions within it, such as distances between dark and bright fringes. It's integral in applying formulas that help unravel interference phenomena.

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Most popular questions from this chapter

35.57. Two thin parallel slits are made in an opaque sheet of film. When a monochromatic beam of light is shone through them at normal incidence, the first bright fringes in the transmitted light occur in air at \(\pm 18.0^{\circ}\) with the original direction of the light beam on a distant screen when the apparatus is in air. When the apparatus is immersed in a liquid, the same bright fringes now occur at \(\pm 12.6^{\circ} .\) Find the index of refraction of the liquid.

35.55. A source \(S\) of monochrounatic light and a detector \(D\) are both located in air a distance \(h\) above a horizontal plane sheet of glass, and are separated by a horizontal distance \(x\) . Waves reaching \(D\) directly from \(S\) interfere with waves that reflect off the glass. The distance \(x\) is small compared to \(h\) so that the reflection is at close to normal incidence. (a) Show that the condition for constructive interference is \(\sqrt{x^{2}+4 h^{2}}-x=\left(m+\frac{1}{2}\right) \lambda,\) and the condition for destructive interference is \(\sqrt{x^{2}+4 h^{2}}-x=m \lambda\) (Hint: Take into account the phase change on reflection.) (b) Let \(h=24 \mathrm{cm}\) and \(x=14 \mathrm{cm} .\) What is the longest wavelength for which there will be constructive interference?

35.17. Two thin parallel slits that are 0.0116 \(\mathrm{mm}\) apart are illuminated by a laser beam of wavelength 585 \(\mathrm{nm}\) . (a) On a very large distant screen, what is the total number of bright fringes (those indicating complete constructive interference), including the central fringe and those on both sides of it? 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 value of \(m ? )\) (b) At what angle, relative to the original direction of the beam, will the fringe that is most distant from the central bright fringe occur?

35.27. What is the thinnest film of a coating with \(n=1.42\) on glass \((n=1.52)\) for which destructive interference of the red component \((650 \mathrm{nm})\) of an incident white light beam in air can take place by reflection?

35.22. GPSTransmission. The GPS (Global Positioning System) satellites are approximately 5.18 \(\mathrm{m}\) across and transmit two low-power signals, one of which is at 1575.42 \(\mathrm{MHz}\) (in the UHF band). In a series of laboratory tests on the satellite, you put two 1575.42-MHz UHF transmitters at opposite ends of the satellite. These broadcast in phase uniformly in all directions. You measure the intensity at points on a circle that is several hundred meters in radius and centered on the satellite. You measure angles on this circle relative to a point that lies along the centerline of the satellite (that is, the perpendicular bisector of a line which extends from one transmitter to the other). At this point on the circle, the measured intensity is \(2.00 \mathrm{W} / \mathrm{m}^{2} .\) (a) At how many other angles in the range \(0^{\circ}<\theta<90^{\circ}\) is the intensity also 2.00 \(\mathrm{W} / \mathrm{m}^{2} ?\) (b) Find the four smallest angles in the range \(0^{\circ}<\theta<90^{\circ}\) for which the intensity is \(2.00 \mathrm{W} / \mathrm{m}^{2} .\) (c) What is the intensity at a point on the circle at an angle of \(4.65^{\circ}\) from the centerline?
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