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Chapter 35: Problem 15

35.15. Coherent light with wavelength 600 \(\mathrm{nm}\) passes through two very narrow slits and the interference pattern is observed on a screen 3.00 \(\mathrm{m}\) from the slits. The first-order bright fringe is at 4.84 \(\mathrm{mm}\) from the center of the central bright fringe. For what wavelength of light will the first order dark fringe be observed at this same point on the screen?

Short Answer

Expert verified
The wavelength for the first-order dark fringe is 1200 nm.

Step by step solution

01

Understand Interference Pattern for Bright Fringes

When light passes through two slits, it creates an interference pattern different from each slit. For bright fringes, the path difference between the two waves must be an integer multiple of the wavelength. Mathematically, this is given by \(d \sin \theta = m \lambda\), where \(d\) is the slit separation, \(\theta\) is the angle to the fringe, \(m\) is the order of the fringe, and \(\lambda\) is the wavelength.
02

Derive Equation from Given Information

We know the first-order bright fringe occurs for wavelength 600 nm at 4.84 mm from the center. For this setup, the equation is \(d \sin \theta = 1 \times 600 \times 10^{-9} \text{ m}\). The distance to the screen \(L = 3.00 \text{ m}\) and the y-distance from the center to the fringe \(y = 4.84 \times 10^{-3} \text{ m}\). Using the small angle approximation, \(\sin \theta \approx \tan \theta \approx \frac{y}{L}\), so \(d \frac{y}{L} = 600 \times 10^{-9} \text{ m}\).
03

Calculate Slit Separation (d)

Though unspecified, we can bypass computing \(d\) directly by using it as a ratio and keeping the equivalent value for \(\sin \theta\). Simplifying gives, \(d = \frac{600 \times 10^{-9} \times 3.00}{4.84 \times 10^{-3}} \text{ m}\). This will allow us to find an equivalent wavelength for the next condition (first-order dark fringe).
04

Condition for First-Order Dark Fringe

For dark fringes, the path difference is given by \(d \sin \theta = (m + \frac{1}{2}) \lambda\). Here, \(m = 0\) for the first-order dark fringe. Plug in \(\sin \theta = \frac{4.84 \times 10^{-3}}{3}\) back: \((\frac{600 \times 10^{-9} \text{ m} \times 3}{4.84 \times 10^{-3} \text{ m}}) \cdot \frac{4.84 \times 10^{-3}}{3} = \frac{1}{2} ~ \lambda_\text{dark}\).
05

Solve for the Wavelength of the First-Order Dark Fringe

Simplify to find \(\lambda_\text{dark}\), therefore \(600 \times 10^{-9} \text{ m} = \frac{1}{2} ~ \lambda_\text{dark}\). Solving \(\lambda_\text{dark} = 1200 \times 10^{-9} \text{ m}\). Convert to nm, \(\lambda_\text{dark} = 1200 \text{ nm}\).

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Key Concepts

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

Coherent Light
Coherent light is a fundamental element in creating interference patterns. For light to be considered coherent, it must have waves that are synchronized, or "in phase," with one another. This means that both the frequency and wavelength are constant across the light waves.
An excellent example of coherent light is laser light. Why coherence matters:
  • It ensures that the phases of the light waves align, allowing for constructive and destructive interference.
  • This alignment is what makes the interference pattern visible.
Light from everyday sources, like light bulbs, isn't generally coherent; the waves are out of phase. However, specific setups are used to make light coherent, such as using double slits. Through these slits, incoming coherent light will create distinctive patterns of bright and dark spots on a screen, demonstrating wave-like properties of light.
Wavelength Determination
Finding the wavelength of light involves understanding the relationship between light interference and geometry. In the exercise, the problem involves determining the wavelength that leads to a first-order dark fringe in a double-slit experiment.Angle approximation:
  • The small angle approximation (\( an \theta \approx \sin \theta \approx \frac{y}{L} \)) simplifies calculations, given the screen is positioned at a distance where angles are small enough that the sine and tangent functions are approximately equal.
  • This approximation allows you to express the angle of light interfered bands effectively using position and distance measures, making theoretical measures approach realistic observations.
By understanding this approximation context, you can insert it into the formula for both bright and dark fringes:
  • For bright fringes: \( d \sin \theta = m \lambda \)
  • For dark fringes: \( d \sin \theta = (m + \frac{1}{2}) \lambda \)
These are key to determining unknown wavelengths just from fringe patterns.
Bright and Dark Fringes
Bright and dark fringes are the telltale signs of light interference. When coherent light passes through double slits, these fringes appear as alternating light and dark bands on a screen. They arise from the constructive and destructive interference of the light waves.Understanding the patterns:
  • Bright Fringes: These occur at locations where the difference in path length between light waves from each slit is a whole number multiple of the light's wavelength. This constructive interference results in brighter spots, mathematically expressed as \(d \sin \theta = m \lambda\), where \(m\) is an integer.
  • Dark Fringes: These occur at locations where the path difference is a half multiple of the wavelength, leading to destructive interference, resulting in darkness. This is captured by the formula \(d \sin \theta = (m + \frac{1}{2}) \lambda\).
By analyzing fringe positions and using these formulas, students can solve problems requiring the determination of wavelengths and further understand wave interference's role in optics.

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Most popular questions from this chapter

35.52. Red light with wavelength 700 \(\mathrm{nm}\) is passed through a two- slit apparatus. At the same time, monochromatic visible light with another wavelength passes through the same apparatus. As a result, most of the pattern that appears on the screen is a mixture of two colors; however, the center of the third bright fringe \((m=3)\) of the red light appears pure red, with none of the other color. What are the possible wavelengths of the second type of visible light? Do you need to know the slit spacing to answer this question? Why or why not?

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 \(\mathrm{nm} ?\)

35.28. Nonglare Glass. When viewing a piece of art that is behind glass, one often is affected by the light that is reflected off the front of the glass (called glare), which can make it difficult to see the art clearly. One solution is to coat the outer surface of the glass with a film to cancel part of the glare. (a) If the glass has a refractive index of 1.62 and you use \(\mathrm{TiO}_{2}\) , which has an index of refraction of \(2.62,\) as the coating, what is the minimum film thickness that will cancel light of wavelength 505 \(\mathrm{nm}\) ? (b) If this coating is too thin to stand up to wear, what other thickness would also work? Find only the three thinnest ones.

35.2. Radio Interference. Two radio antennas \(A\) and \(B\) radiate in phase. Antenna \(B\) is 120 \(\mathrm{m}\) to the right of antenna \(A\) . Consider point \(Q\) along the extension of the line connecting the antennas, a horizontal distance of 40 \(\mathrm{m}\) to the right of antenna \(B\) . The frequency, and hence the wavelength, of the emitted waves can be varied. (a) What is the longest wavelength for which there will be destructive interference at point \(Q ?\) (b) What is the longest wave-length for which there will be constructive interference at point \(Q\) ?

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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