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Chapter 27: Problem 6

Two parallel slits are illuminated by light composed of two wavelengths. One wavelength is \(\lambda_{\mathrm{A}}=645 \mathrm{nm} .\) The other wavelength is \(\lambda_{\mathrm{B}}\) and is unknown. On a viewing screen, the light with wavelength \(\lambda_{\mathrm{A}}=645 \mathrm{nm}\) produces its third-order bright fringe at the same place where the light with wavelength \(\lambda_{\mathrm{B}}\) produces its fourth dark fringe. The fringes are counted relative to the central or zeroth-order bright fringe. What is the unknown wavelength?

Short Answer

Expert verified
The unknown wavelength is 430 nm.

Step by step solution

01

Understand the Fringe Conditions

For the bright fringes produced by light with wavelength \(\lambda_A\), the condition is \(d \sin \theta = m \lambda_A\), where \(m\) is the order of the bright fringe. For the dark fringes produced by light with wavelength \(\lambda_B\), the condition is \(d \sin \theta = (n+1/2)\lambda_B\), where \(n\) is the order of the dark fringe. We know that the third bright fringe for wavelength \(\lambda_A\) coincides with the fourth dark fringe for wavelength \(\lambda_B\).
02

Set Up the Equations

For the third-order bright fringe of \(\lambda_A\), we have:\[d \sin \theta = 3 \lambda_A\]For the fourth-order dark fringe of \(\lambda_B\), we have:\[d \sin \theta = 4.5 \lambda_B\]Since these two distances are equal, set the equations equal to each other:\[3 \lambda_A = 4.5 \lambda_B\]
03

Solve for the Unknown Wavelength

Use the equation from Step 2 to find the unknown wavelength \(\lambda_B\):\[3 \lambda_A = 4.5 \lambda_B\]\[\lambda_B = \frac{3 \lambda_A}{4.5}\]Substitute \(\lambda_A = 645 \mathrm{nm}\):\[\lambda_B = \frac{3 \times 645}{4.5}\]\[\lambda_B = 430 \mathrm{nm}\]

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Key Concepts

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

Interference Patterns
Interference patterns are a fascinating phenomenon that arise when two or more waves overlap and combine with each other. This concept is foundational in understanding the double slit experiment, a classic demonstration in physics. The patterns arise because of the constructive and destructive interference of light waves.
  • Constructive Interference: This occurs when the waves align perfectly, meaning their peaks and troughs match up. This results in a brighter, more intense light spot, known as a bright fringe.
  • Destructive Interference: On the other hand, when the waves are out of phase 鈥� peaks align with troughs 鈥� the waves effectively cancel each other out, creating a dark fringe.
By illuminating two slits with light and examining the pattern on a screen, we can observe the alternating series of bright and dark fringes. These patterns depend on the wavelength of the light used and the geometry of the setup, such as the distance between the slits and the screen. Understanding these interference patterns helps in exploring wave properties of light and similar phenomena in other wave types.
Order of Fringes
The order of fringes relates to the sequence of bright and dark spots seen in an interference pattern. Each bright or dark spot is labeled with an order number, typically starting with zero at the central maximum.
  • In the scenario of a double slit experiment, **bright fringes** are numbered by an integer (m), starting from the central maximum (zeroth order).
  • **Dark fringes**, representing positions of destructive interference, generally have non-integer indices like half-integers ((n + 1/2){\lambda}_B).
These order numbers help specify which bright or dark fringe we are talking about. For instance, in the given exercise, the third bright fringe for a wavelength \(\lambda_A\) and the fourth dark fringe for \(\lambda_B\) coincide, which is significant for wavelength calculations. Using the order of fringes, one can better understand how different wavelengths produce specific interference patterns, which can be crucial for calculating unknown wavelengths.
Wavelength Calculation
Wavelength calculation is an essential aspect of analyzing interference patterns. In many cases, as with the textbook problem, you might need to determine an unknown wavelength based on the interference pattern it produces relative to another known wavelength. To calculate the unknown wavelength, one must use the relationships established for bright and dark fringes.
  • The formula for **bright fringes** is \(d \sin \theta = m \lambda\), where "\(m\)" is the fringe order.
  • For **dark fringes**, the formula is \(d \sin \theta = (n + 1/2) \lambda\).
By equating these when fringes coincide, as was done in the problem, one can solve for the unknown wavelength. This involves setting the equations for two coinciding fringes equal, rearranging the formula to isolate the unknown wavelength, and substituting known values. In the example, substituting the known \(\lambda_A = 645 \mathrm{nm}\) in the formula allowed calculating \(\lambda_B = 430 \mathrm{nm}\). Calculations like these illustrate the precise and mathematical nature of understanding light behavior in wave optics.

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Most popular questions from this chapter

You are standing in air and are looking at a flat piece of glass \((n=\) 1.52 ) on which there is a layer of transparent plastic \((n=1.61) .\) Light whose wavelength is 589 nm in vacuum is incident nearly perpendicularly on the coated glass and reflects into your eyes. The layer of plastic looks dark. Find the two smallest possible nonzero values for the thickness of the layer.

Crude Thickness Monitor. You and your team need to determine the thicknesses of two extremely thin sheets of plastic and do not have a measurement instrument capable of the job. You find a white light source with color filters (i.e., tinted glass windows that only let certain colors pass through) including red \((\lambda=660 \mathrm{nm}),\) green \((\lambda=550 \mathrm{nm}),\) and blue \((\lambda=\) \(470 \mathrm{nm}) .\) You locate two glass microscope slides, each of length \(10.0 \mathrm{cm}\) stack one on top of the other, and tape their ends together on one side. On the very edge of the side opposite the tape, you sandwich a piece of plastic between the slides to form a wedge of air between the plates. (a) When you shine blue light perpendicular to the slides, 55 bright fringes form along the full length of the slide. How thick is the plastic? (b) When you replace the first sheet with the second sheet of plastic, the fringes that appear are too closely spaced to count. You switch to red light (a longer wavelength), and count 124 fringes. How thick is the plastic? (c) How many fringes would you expect in each case, (a) and (b), if you instead used green light?

At most, how many bright fringes can be formed on either side of the central bright fringe when light of wavelength \(625 \mathrm{nm}\) falls on a double slit whose slit separation is \(3.76 \times 10^{-6} \mathrm{m} ?\)

In a single-slit diffraction pattern, the central fringe is 450 times as wide as the slit. The screen is 18000 times farther from the slit than the slit is wide. What is the ratio \(\lambda / W,\) where \(\lambda\) is the wavelength of the light shining through the slit and \(W\) is the width of the slit? Assume that the angle that locates a dark fringe on the screen is small, so that \(\sin \theta \approx \tan \theta\).

An Optical Monochromator. You and your team are designing a device that inputs a beam of white light (i.e., a continuous spectrum of visible light spanning all wavelengths from \(410 \mathrm{nm}\) to \(660 \mathrm{nm}\) ), and outputs a nearly monochromatic beam (i.e., a single color). Such a device is called an optical monochromator and is used in a wide variety of instruments and scientific experiments. In the instrument you are building, white light impinges upon the backside of a diffraction grating that has 1200 lines/cm. A movable rectangular aperture (a slit) is located on the opposite side of the grating, and can translate along a circular arc of radius \(20.0 \mathrm{cm},\) the center of which is located at the grating. (a) At what angle relative to the normal of the grating should the center of the slit be located in order to pass green light \((\lambda=550 \mathrm{nm})\) from first order \((m=1)\) diffracted light? (b) How wide should the slit be so that the wavelengths passing through the slit fall in the range \(540 \mathrm{nm} \leq \lambda \leq 560 \mathrm{nm} ?\)
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