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Chapter 35: Problem 13

Two very narrow slits are spaced 1.80 \(\mu\)m apart and are placed 35.0 cm from a screen. What is the distance between the first and second dark lines of the interference pattern when the slits are illuminated with coherent light with \(\lambda\) = 550 nm? (Hint: The angle \(\theta\) in Eq. (35.5) is \(not\) small.)

Short Answer

Expert verified
The distance between the first and second dark lines is approximately 3.1 mm.

Step by step solution

01

Understand the Problem

We need to calculate the distance between the first and second dark lines in the interference pattern from two slits using the given wavelength \(\lambda = 550\ \text{nm}\), slit separation \(d = 1.80\ \mu\text{m}\), and distance to the screen \(L = 35\ \text{cm}\). The dark lines appear where destructive interference occurs, given by the path difference \(d\sin\theta = \left(m + \frac{1}{2}\right)\lambda\), where \(m\) is an integer for dark lines.
02

Use the Condition for Dark Lines

For the first dark line, \(m = 0\) so the condition becomes:\[d\sin\theta_1 = \left(0 + \frac{1}{2}\right)\lambda = \frac{1}{2}\lambda\]For the second dark line, \(m = 1\) so:\[d\sin\theta_2 = \left(1 + \frac{1}{2}\right)\lambda = \frac{3}{2}\lambda\]
03

Calculate the Angles \(\theta_1\) and \(\theta_2\)

Solve for \(\theta_1\) and \(\theta_2\) using the equations:\[\sin\theta_1 = \frac{\frac{1}{2}\lambda}{d} = \frac{275}{1800}\approx 0.1528\]\[\sin\theta_2 = \frac{\frac{3}{2}\lambda}{d} = \frac{825}{1800}\approx 0.4583\]Find \(\theta_1 = \arcsin(0.1528)\) and \(\theta_2 = \arcsin(0.4583)\).
04

Calculate Position of Dark Lines on the Screen

Use \(y = L\tan\theta\) to find the position \(y_1\) for the first dark line and \(y_2\) for the second:\[y_1 = 35\times\tan(\theta_1)\]\[y_2 = 35\times\tan(\theta_2)\]Compute \(y_1\) and \(y_2\).
05

Find the Distance Between the Dark Lines

The distance \(\Delta y\) between the first and second dark lines is:\[\Delta y = y_2 - y_1\]Calculate the difference between the two positions obtained in Step 4 to find \(\Delta y\).

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Key Concepts

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

Destructive interference
When we talk about interference patterns, "destructive interference" is a crucial concept. This happens when two or more waves overlap in such a way that they cancel each other out. The high points of one wave meet the low points of another, resulting in reduced or zero amplitude. This effect is very easy to observe in light waves, with the result being areas that appear dark on a screen.

In the context of our double-slit experiment, destructive interference results in dark lines on the screen where the waves from the two slits perfectly cancel each other out. This occurs when the path difference between the two waves is a half-multiple of the wavelength, specifically \[ d\sin\theta = \left(m + \frac{1}{2}\right)\lambda \]where:
  • \(d\) is the distance between the slits,
  • \(\lambda\) is the wavelength of the light,
  • \(\theta\) is the angle of the dark fringe from the horizontal,
  • \(m\) is an integer (0, 1, 2, ...).
This formula helps us find where the darkness happens, indicating complete wave cancellation on our pattern.
Double-slit experiment
The double-slit experiment, first conducted by Thomas Young, is a famous demonstration of the wave properties of light. In this experiment, coherent light 鈥� light of a single wavelength 鈥� shines on a pair of closely-spaced slits. As the light passes through these slits, it spreads out or diffracts.

The resulting light waves overlap and interfere with each other, producing distinct patterns on a screen behind the slits. We see a series of bright and dark lines. Bright lines occur from constructive interference, while dark lines, where no light appears, result from destructive interference.

This experiment is not only foundational in understanding wave properties of light but also delves into the principles of superposition, where the resulting wave pattern is a sum of individual wave patterns. As in our problem, by knowing slit spacing and wavelength, interference patterns teach us about wave interactions on a detailed level.
Wavelength calculation
Wavelength calculation is an integral part of solving interference problems. Wavelength, denoted as \(\lambda\), measures the distance between consecutive peaks of a wave. It's a key parameter that determines where the interference pattern's bright and dark spots will appear.

In calculating where dark lines form in interference patterns (as in our example), we use the formula specific to destructive interference:\[ d\sin\theta = \left(m + \frac{1}{2}\right)\lambda \]Here, \(m\) is the order of the dark fringe, and knowing the position \(\theta\) helps us locate these dark spots precisely on the screen.

To find where these lines appear on the screen, converting angles to linear position using \( y = L\tan\theta \), where \(L\) is the distance from the slits to the screen, is crucial. All these calculations revolve around understanding wavelength, showing its central role in predicting wave interference patterns effectively.

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Most popular questions from this chapter

After a laser beam passes through two thin parallel slits, the first completely dark fringes occur at \(\pm\)19.0\(^\circ\) with the original direction of the beam, as viewed on a screen far from the slits. (a) What is the ratio of the distance between the slits to the wavelength of the light illuminating the slits? (b) What is the smallest angle, relative to the original direction of the laser beam, at which the intensity of the light is \(1 \over 10\) the maximum intensity on the screen?

The index of refraction of a glass rod is 1.48 at \(T\) =20.0\(^\circ\)C and varies linearly with temperature, with a coefficient of 2.50 \(\times\) 10\(^{-5}\)/C\(^\circ\). The coefficient of linear expansion of the glass is 5.00 \(\times\) 10\(^{-6}\)/C\(^\circ\). At 20.0\(^\circ\)C the length of the rod is 3.00 cm. A Michelson interferometer has this glass rod in one arm, and the rod is being heated so that its temperature increases at a rate of 5.00 C\(^\circ\)/min. The light source has wavelength \(\lambda\) = 589 nm, and the rod initially is at \(T\) = 20.0\(^\circ\)C. How many fringes cross the field of view each minute?

Two speakers \(A\) and \(B\) are 3.50 m apart, and each one is emitting a frequency of 444 Hz. However, because of signal delays in the cables, speaker \(A\) is one-fourth of a period ahead of speaker \(B\). For points far from the speakers, find all the angles relative to the centerline (Fig. P35.44) at which the sound from these speakers cancels. Include angles on both sides of the centerline. The speed of sound is 340 m/s.

Laser light of wavelength 510 nm is traveling in air and shines at normal incidence onto the flat end of a transparent plastic rod that has \(n\) = 1.30. The end of the rod has a thin coating of a transparent material that has refractive index 1.65. What is the minimum (nonzero) thickness of the coating (a) for which there is maximum transmission of the light into the rod; (b) for which transmission into the rod is minimized?

Two slits spaced 0.450 mm apart are placed 75.0 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 nm?
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