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Chapter 35: Problem 16

Coherent light that contains two wavelengths, 660 \(\mathrm{nm}\) (red) and 470 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?

Short Answer

Expert verified
The distance is 3.17 mm between the first-order bright fringes for the red and blue wavelengths.

Step by step solution

01

Identify Key Variables and Formula

First, note the given wavelengths: 660 nm (red) and 470 nm (blue), as well as the slit separation \(d = 0.300\, \text{mm} = 0.0003\, \text{m}\) and the distance to the screen \(L = 5.00\, \text{m}\). The formula for the position \(y\) of the m-th order bright fringe is \(y = \frac{m \lambda L}{d}\), where \(m\) is the order (1 in this case).
02

Calculate Position for Red Fringe (660 nm)

Substitute the red wavelength into the formula: \(\lambda = 660 \times 10^{-9}\, \text{m}\), \(m = 1\), \(L = 5.00\, \text{m}\), \(d = 0.0003\, \text{m}\). So, \(y_\text{red} = \frac{660 \times 10^{-9} \times 5.00}{0.0003}\).
03

Solve for Position of Red Fringe

Calculate \(y_\text{red}\):\[y_\text{red} = \frac{660 \times 10^{-9} \times 5.00}{0.0003} = 0.011\, \text{m} = 11.0\, \text{mm}.\]
04

Calculate Position for Blue Fringe (470 nm)

Use \(\lambda = 470 \times 10^{-9}\, \text{m}\). Substitute into the formula: \(y_\text{blue} = \frac{470 \times 10^{-9} \times 5.00}{0.0003}\).
05

Solve for Position of Blue Fringe

Calculate \(y_\text{blue}\):\[y_\text{blue} = \frac{470 \times 10^{-9} \times 5.00}{0.0003} = 0.00783\, \text{m} = 7.83\, \text{mm}.\]
06

Find Distance Between First Order Fringes

The distance between the first-order bright fringes for the two wavelengths is the difference: \(\Delta y = y_\text{red} - y_\text{blue}\).
07

Calculate Delta y

Calculate the distance: \[\Delta y = 11.0\, \text{mm} - 7.83\, \text{mm} = 3.17\, \text{mm}.\]

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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 type of light where the waves have a constant phase difference, the same frequency, and, often, the same amplitude. These properties make coherent light essential for producing clear and stable interference patterns, such as those seen in physics experiments involving light and slits. In most cases, coherent light is achieved using lasers, due to their precise control over these properties.
  • Constant Phase Difference: This means that at any given point, the phase difference between two light waves remains unchanged over time.
  • Same Frequency: Coherent waves also have the same frequency, meaning the number of oscillations or vibrations per second is identical.
  • Same Amplitude: While not always necessary, having the same amplitude helps in achieving maximum interference effects.
Coherent light is crucial for experiments like the one described, where precise measurements of interference patterns are required.
Wavelength
Wavelength is the distance between consecutive peaks of a wave, often represented by the symbol \(\lambda\). It is a fundamental characteristic of all waves, including light, and determines the color of visible light. In the exercise, two different wavelengths are used: 660 nm (red light) and 470 nm (blue light).
  • Red Light (660 nm): Longer wavelength, and is less refracted through different mediums, which often makes red light appear more spread out in an interference pattern.
  • Blue Light (470 nm): Shorter wavelength, which results in higher frequencies and often more pronounced bending and scattering.
The difference in these wavelengths is what allows us to see distinct positions of their corresponding bright fringes on the screen.
Slit Separation
Slit separation, denoted as \(d\), is the distance between the two slits through which the light passes in a double-slit experiment. This distance is a critical factor in determining the nature and spacing of the interference pattern created. In this exercise, the slit separation is 0.300 mm.The role of slit separation is vital because:
  • The smaller the slit separation, the wider the fringes will appear on the screen.
  • Conversely, increasing the slit separation will cause the fringes to be closer together.
By adjusting the slit separation, one can control the layout of the interference pattern, a principle commonly applied in various optical devices and tests.
Bright Fringes
Bright fringes, which appear in interference patterns, are areas of constructive interference where two or more light waves superimpose each other and the amplitudes add up to create a brighter effect. These fringes are the result of coherent light waves meeting in phase. Their precise positions are determined by multiple factors:
  • Wavelength: Different wavelengths will produce bright fringes at different positions due to varying distances between wave peaks.
  • Order of Interference: In the given problem, we consider the first-order (\(m=1\)) bright fringe, which is the closest to the central maximum.
  • Slit Separation: As mentioned, the wider the slits are apart, the closer the bright fringes will be.
In practical terms, bright fringes are where you observe the most noticeable and vivid light bands, crucial for analyzing wave behaviors.

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Most popular questions from this chapter

A plastic film with index of refraction 1.85 is put on the surface of a car window to increase the reflectivity and thus to keep the interior of the car cooler. The window glass has index of refraction \(1.52 .\) (a) What minimum thickness is required if light with wavelength 550 \(\mathrm{nm}\) in air reflected from the two sides of the film is to interfere constructively? (b) It is found to be difficult to manufacture and install coatings as thin as calculated in part (a). What is the next greatest thickness for which there will also be constructive interference?

A source \(S\) of monochromatic 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?

Two speakers that are 15.0 \(\mathrm{m}\) apart produce in-phase sound waves of frequency 250.0 \(\mathrm{Hz}\) in a room where the speed of sound is 340.0 \(\mathrm{m} / \mathrm{s} .\) A woman starts out at the midpoint between the two speakers. The room's walls and ceiling are covered with absorbers to eliminate reflections, and she listens with only one ear for best precision. (a) What does she hear: constructive or destructive interference? Why? (b) She now walks slowly toward one of the speakers. How far from the center must she walk before she first hears the sound reach a minimum intensity? (c) How far from the center must she walk before she first hears the sound maximally enhanced?

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

Coherent light of frequency \(6.32 \times 10^{14} \mathrm{Hz}\) passes through two thin slits and falls on a screen 85.0 \(\mathrm{cm}\) away. You observe that the third bright fringe occurs at \(\pm 3.11 \mathrm{cm}\) on either side of the central bright fringe. (a) How far apart are the two slits? (b) At what distance from the central bright fringe will the third dark fringe occur?
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