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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
The distance between the second and third dark lines is approximately 0.0835 cm.

Step by step solution

01

Understand the Problem

We're dealing with a classic double-slit experiment. Here, the task is to find the distance between the second and third dark lines (minima) of the interference pattern on the screen. We know the slit separation, the wavelength of the light, and the distance to the screen.
02

Calculate the Angle for Dark Lines

The formula for the angular position of the minima (dark lines) is: \( d \sin\theta = (m + \frac{1}{2})\lambda \), where \( m \) is the order of minima, \( d \) is the slit separation, \( \lambda \) is the wavelength. Since the dark lines required are the second and third, it means \( m_2 = 1 \) (for the second dark line) and \( m_3 = 2 \) (for the third dark line).
03

Determine the Angle for Each Minima

For the second minima: \( 0.450 \times 10^{-3} \sin\theta_2 = (1.5) \times 500 \times 10^{-9} \). Solve for \( \theta_2 \).For the third minima: \( 0.450 \times 10^{-3} \sin\theta_3 = (2.5) \times 500 \times 10^{-9} \). Solve for \( \theta_3 \).
04

Solving for \( \theta \)

For the second minima, \( \sin\theta_2 = \frac{750 \times 10^{-9}}{0.450 \times 10^{-3}} = 1.6667 \times 10^{-3} \), resulting in \( \theta_2 \approx 0.0954^{\circ} \). For the third minima, \( \sin\theta_3 = \frac{1250 \times 10^{-9}}{0.450 \times 10^{-3}} = 2.7778 \times 10^{-3} \), resulting in \( \theta_3 \approx 0.1590^{\circ} \).
05

Calculate the Distance Between the Minima on the Screen

The distance \( y \) from the center to a minima is given by \( y = L \tan\theta \). Find \( y_2 = 75 \times \tan(0.0954^{\circ}) \) and \( y_3 = 75 \times \tan(0.1590^{\circ}) \). Then, calculate the distance between these two dark lines as \( \Delta y = y_3 - y_2 \).
06

Compute the Final Answer

Calculate: \( y_2 \approx 75 \times 0.001665 = 0.12495 \text{ cm} \) and \( y_3 \approx 75 \times 0.002779 = 0.208425 \text{ cm} \). Thus, \( \Delta y = 0.208425 - 0.12495 = 0.083475 \text{ cm} \).

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Key Concepts

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

Interference Pattern
The double-slit experiment is renowned for demonstrating wave interference. When coherent light passes through two closely spaced slits, it creates an interference pattern on a screen placed at some distance. This pattern consists of alternating bright and dark bands. The bright bands are lines of constructive interference where the waves from the two slits arrive in phase, reinforcing each other. The dark bands, on the other hand, are lines of destructive interference where the waves cancel each other out. This occurs when their phase difference results in a complete cancellation.
The distance between these alternating bands can tell us a great deal about the properties of the light used, such as its wavelength and the slit separation. In the case of the original exercise, understanding the distribution of these lines helps us find the distance between specific dark lines, which are essentially points of no light on the screen.
Dark Lines Minima
Dark lines, or minima, in an interference pattern are regions where destructive interference occurs. In simpler terms, where light waves from the two slits meet in opposite phases, causing them to cancel out. This results in a dark spot on the screen. The formula used to find these dark lines is:
  • \( d \sin\theta = (m + \frac{1}{2})\lambda \)
where:
  • \(d\) is the distance between the two slits,
  • \(\theta\) is the angle at which a dark line appears,
  • \(m\) is the order of the minimum and can be an integer (0, 1, 2,...),
  • \(\lambda\) is the wavelength of the light.
For example, in the case of the exercise, the second dark line corresponds to \(m = 1\) and the third to \(m = 2\). Calculating \(\theta\) for these values allows us to determine their position on the screen.
Coherent Light
Coherent light is a type of light where the waves are synchronized. This means they have a constant phase difference and generally the same frequency and wavelength. Lasers are a common source of coherent light, often used in experiments like the double-slit because their monochromatic and coherent nature creates clear interference patterns.
The coherence of the light ensures that when it goes through the slits, it spreads out into waves that can consistently interfere - either constructively or destructively. Without coherence, the interference pattern would be muddled and unclear. In the exercise example, using coherent light with a wavelength of 500 nm ensures precise and predictable dark and bright lines.
Wavelength Calculation
Calculating the wavelength allows us to explore many aspects of wave behavior. In a double-slit experiment, the wavelength significantly influences the spacing of the interference pattern's bands. Once you know the slit separation and can observe the interference pattern, you can reverse-engineer to find the wavelength using the formula:
  • \( \lambda = \frac{d \sin\theta}{m + \frac{1}{2}} \)
This ability lets scientists and engineers make sense of the light's properties by looking at how it interacts with the slits.
In the exercise's context, given that the distance between the dark lines was already calculated using a known wavelength, starting from the interference pattern could conversely enable a calculation of \(\lambda\) if needed. By understanding the positions of dark minima and the angles they are at, the inverse process helps uncover the light’s unique characteristics.

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Most popular questions from this chapter

Coherent light that contains two wavelengths, \(660 \mathrm{nm}\) (red) and \(470 \mathrm{nm}\) (blue), passes through two narrow slits separated by \(0.300 \mathrm{~mm},\) and the interference pattern is observed on a screen \(5.00 \mathrm{~m}\) from the slits. What is the distance on the screen between the first-order bright fringes for the two wavelengths?

Due to blurring caused by atmospheric distortion, the best resolution that can be obtained by a normal, earth-based, visible-light telescope is about 0.3 arcsecond (there are 60 arcminutes in a degree and 60 arcseconds in an arcminute). (a) Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with \(550 \mathrm{nm}\) light. (b) Increasing the telescope diameter beyond the value found in part (a) will increase the light-gathering power of the telescope, allowing more distant and dimmer astronomical objects to be studied, but it will not improve the resolution. In what ways are the Keck telescopes (each of \(10 \mathrm{~m}\) diameter) atop Mauna Kea in Hawaii superior to the Hale Telescope ( 5 m diameter) on Palomar Mountain in California? In what ways are they not superior? Explain.

Monochromatic light from a distant source is incident on a slit \(0.750 \mathrm{~mm}\) wide. On a screen \(2.00 \mathrm{~m}\) away, the distance from the central maximum of the diffraction pattern to the first minimum is measured to be \(1.35 \mathrm{~mm}\). Calculate the wavelength of the light.

Light of wavelength \(631 \mathrm{nm}\) passes through a diffraction grating having 485 lines \(/ \mathrm{mm}\). (a) What is the total number of bright spots (indicating complete constructive interference) that will occur on a large distant screen? Solve this problem without finding the angles. (Hint: What is the largest that \(\sin \theta\) can be? What does this imply for the largest value of \(m ?\) (b) What is the angle of the bright spot farthest from the center?

Eyeglass lenses can be coated on the inner surfaces to reduce the reflection of stray light to the eye. If the lenses are medium flint glass of refractive index 1.62 and the coating is fluorite of refractive index \(1.432,\) (a) what minimum thickness of film is needed on the lenses to cancel light of wavelength \(550 \mathrm{nm}\) reflected toward the eye at normal incidence, and (b) will any other wavelengths of visible light be canceled or enhanced in the reflected light?

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