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Chapter 17: Problem 10

Figure \(P 17.10\) shows the fringes observed in a doubleslit interference experiment nated by white light. The central maximum is white because all of the colors overlap. This is not true for the other fringes. The \(m=1\) fringe clearly shows bands of color, with red appearing farther from the center of the pattern, and blue closer. If the slits that create this pattern are \(20 \mu \mathrm{m}\) apart and are located \(0.85 \mathrm{m}\) from the screen, what are the \(m=1\) distances from the central maximum for red \((700 \mathrm{nm})\) and violet \((400 \mathrm{nm})\) light?

Short Answer

Expert verified
The fringe distances from the central maximum for red and violet light can be obtained by calculating values from Steps 2 and 3 respectively.

Step by step solution

01

Convert Units of Given Values

First, it's important to maintain consistency in the measurement unit for the calculation. Convert the given values to meters: slit-to-screen distance \(L = 0.85 m\), slit distance \(d = 20 \mu m = 20 \times 10^{-6} m\), the wavelength of red light \(\lambda = 700 nm = 700 \times 10^{-9} m\) and the wavelength of violet light \(\lambda = 400 nm = 400 \times 10^{-9} m\).
02

Calculate Fringe Spacing for Red Light

Substitute the given values into the fringe spacing formula: \(\Delta y_{red} = \frac{m\lambda L}{d} = \frac{1 \times (700 \times 10^{-9}) \times 0.85}{20 \times 10^{-6}}\). Calculate the value to find the fringe distance for red light.
03

Calculate Fringe Spacing for Violet Light

Again, substitute the given values into the formula, this time with the wavelength for violet light: \(\Delta y_{violet} = \frac{m\lambda L}{d} = \frac{1 \times (400 \times 10^{-9}) \times 0.85}{20 \times 10^{-6}}\). Calculate the value to find the fringe distance for violet light.

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Key Concepts

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

Fringe Spacing Calculation
Understanding the concept of fringe spacing is crucial when analyzing patterns in a double-slit interference experiment. Fringe spacing, also known as fringe width or interfringe distance, is the separation between consecutive bright (or dark) bands on a screen caused by the constructive (or destructive) interference of light.

The formula used to calculate the fringe spacing in a double-slit experiment is \(\frac{m\theta L}{d}\). Here, \(m\) is the order of the fringe, \(L\) is the distance from the slits to the screen, \(d\) is the distance between the slits, and \(\theta\) is the wavelength of the light used. The fringe spacing is directly proportional to both the wavelength and the distance from the slits to the screen. It is inversely proportional to the distance between the slits.

Applying this formula helps us understand how different colors of light, which have different wavelengths, will produce different fringe spacings. This is why, in the given exercise, red light with a longer wavelength creates a larger fringe spacing than violet light with a shorter wavelength.
Wavelength of Light
The wavelength of light is a fundamental property that determines many behaviors of light waves, including their interactions in experiments such as double-slit interference. Wavelength is the distance between successive points of equal phase in the wave, such as crest to crest or trough to trough, and is usually denoted by the symbol \(\theta\).

In the visible spectrum, wavelength ranges from about 400 nanometers (violet) to 700 nanometers (red). The variation in wavelength is what gives light its color when it comes to visible light. This property is exploited in a double-slit experiment to demonstrate how light of different colors will generate patterns of varying dimensions due to their respective wavelengths.

It's worth noting that when converting wavelengths for calculations, consistency in units is paramount. Typically, wavelengths are measured in meters for mathematical operations, requiring the conversion of nanometers to meters by multiplying by \(10^{-9}\). This ensures accuracy in the calculation of physical quantities like fringe spacing.
Constructive and Destructive Interference
Interference of light waves is a phenomenon that occurs when two or more waves overlap and combine to form a new wave pattern. This effect is central to the double-slit interference experiment and can be categorized into two types: constructive interference and destructive interference.

Constructive Interference

Constructive interference happens when waves from different slits arrive at a point in phase 鈥� crest meeting crest and trough meeting trough. This results in a bright fringe or band on the screen as the wave amplitudes reinforce each other, producing a wave with a larger amplitude.

Destructive Interference

Contrastingly, destructive interference occurs when the waves arrive out of phase 鈥� a crest meets a trough, and vice versa. This situation leads to a dark fringe or band because the wave amplitudes cancel each other out, resulting in a wave with reduced or no amplitude.

The bright and dark fringes observed in the double-slit experiment are the direct consequences of constructive and destructive interference. The pattern of these fringes can be predicted and measured, providing valuable insights into the properties of the light being used, such as its wavelength.

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Most popular questions from this chapter

A physics instructor wants to project a spectrum of visiblelight colors from \(400 \mathrm{nm}\) to \(700 \mathrm{nm}\) as part of a classroom demonstration. She shines a beam of white light through a diffraction grating that has 500 lines per \(\mathrm{mm},\) projecting a pattern on a screen \(2.4 \mathrm{m}\) behind the grating. a. How wide is the spectrum that corresponds to \(m=1 ?\) b. How much distance separates the end of the \(m=1\) spectrum and the start of the \(m=2\) spectrum?

Quality control systems have been developed to remotely measure the diameter of wires using diffraction. A wire with a stated diameter of \(170 \mu \mathrm{m}\) blocks the beam of a \(633 \mathrm{nm}\) laser, producing a diffraction pattern on a screen \(50.0 \mathrm{cm}\) distant. The width of the central maximum is measured to be \(3.77 \mathrm{mm}\). The wire should have a diameter within \(1 \%\) of the stated value. Does this wire pass the test?

Diffraction can be used to provide a quick test of the size of red blood cells. Blood is smeared onto a slide, and a laser shines through the slide. The size of the cells is very consistent, so the multiple diffraction patterns overlap and produce an overall pattern that is similar to what a single cell would produce. Ideally, the diameter of a red blood cell should be between 7.5 and \(8.0 \mu \mathrm{m} .\) If a \(633 \mathrm{nm}\) laser shines through a slide and produces a pattern on a screen \(24.0 \mathrm{cm}\) distant, what range of sizes of the central maximum should be expected? Values outside this range might indicate a health concern and warrant further study.

Early investigators (including Thomas Young) measured the thickness of wool fibers using diffraction. One early instrument used a collimated beam of 560 nm light to produce a diffraction pattern on a screen placed \(30 \mathrm{cm}\) from a single wool fiber. If the fiber's diameter was \(16 \mu \mathrm{m},\) what was the width of the central maximum on the screen?

Two narrow slits are illuminated by light of wavelength \(\lambda\). The slits are spaced 20 wavelengths apart. What is the angle, in radians, between the central maximum and the \(m=1\) bright fringe?
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