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Chapter 17: Problem 7

Two narrow slits \(50 \mu \mathrm{m}\) apart are illuminated with light of wavelength \(500 \mathrm{nm} .\) The light shines on a screen \(1.2 \mathrm{m}\) distant. What is the angle of the \(m=2\) bright fringe? How far is this fringe from the center of the pattern?

Short Answer

Expert verified
The angle of the \(m=2\) bright fringe is given by the equation in Step 2. Once the angle is known, you can calculate the distance of this fringe from the center of the pattern using the solution outlined in Step 3.

Step by step solution

01

Set up the formula for the fringe angle

The formula that gives the angle of the bright fringe in a double-slit experiment is described as \(\theta_m = m \cdot \sin^{-1}\Big( \frac{m \lambda}{d} \Big)\) where \(m = \) fringe order \( (2)\), \(\lambda = \) wavelength of light \( (500 nm)\), and \(d = \) distance between the slits \( (50 \mu m)\).
02

Find the angle of the bright fringe

Enter the given values into the formula to find the fringe angle: \(\theta_2 = \sin^{-1}\Big( \frac{2 \cdot 500 \times 10^{-9} m}{50 \times 10^{-6} m} \Big)\). This should give you the angle in radians. Convert it into degrees by multiplying the result by \( \frac{180}{\pi} \).
03

Calculate the distance of the fringe from the center

The distance of the fringe from the center is given by the formula \(x_m = L \tan(\theta_m)\) where \(L = \) distance from the slit to the screen \( (1.2 m)\). Plugging in the calculated value of \(\theta_m\) from Step 2 will give the distance of the fringe from the center.

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Key Concepts

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

Interference Pattern
When we talk about the mysteries of light, one of the most fascinating phenomena to be observed is the interference pattern. This occurs when waves from multiple sources, such as two slits in the case of the famous double-slit experiment, converge and interact with one another.

As these light waves, which have crests and troughs just like ocean waves, meet, they can either reinforce each other (constructive interference) or cancel each other out (destructive interference). Imagine two waves colliding: if the crest of one meets the crest of the other, they build up to create a brighter light at that point. On the other hand, if a crest meets a trough, they tend to nullify each other, leading to darkness. This interaction produces a series of light and dark bands known as an interference pattern, observable on a screen placed behind the slits.

The pattern itself is a direct consequence of the wave-like nature of light, and it鈥檚 crucial for students to comprehend that the spacing and arrangement of these bands give insights into the properties of the light, including its wavelength and the separation of the slits from which it diffracted.
Fringe Angle Calculation
Understanding how to calculate the fringe angle in a double-slit experiment is essential for predicting where the bright and dark bands will appear on the screen. The angle is essentially a measure of how much the peak of a particular band, or 'fringe', is deviated from the central axis. From the formula \( \theta_m = m \cdot \sin^{-1}\Big( \frac{m \lambda}{d} \Big) \), it is evident that the fringe angle depends on the order of the fringe (represented by \(m\)), the wavelength of the light (\(\lambda\)), and the separation of the slits (\(d\)).

For instance, \(m=2\) refers to the second-order bright fringe, which is the second band of light from the center on either side. By plugging in the known values and calculating, students can find out the precise angle at which this second-order bright fringe will appear. This step is crucial as it sets the groundwork for predicting the location of the fringes on the screen.
Wavelength and Slit Separation
The double-slit experiment is not only an excellent demonstration of light's wave properties but also provides a method to measure the light's wavelength and understand the effect of slit separation on the interference pattern. The wavelength (\(\lambda\)) is the distance between successive peaks of a wave and a critical factor in determining the spacing of the interference fringes.

Slit separation (\(d\)), on the other hand, refers to the distance between the two narrow slits through which the light passes. This separation also influences the pattern. If the slits are closer together, the fringes will be more spread out; if the slits are farther apart, the fringes will be closer together. It's an inverse relationship that can sometimes confuse students, but it can be mastered with practice.

Students should note that the combination of wavelength and slit separation not only determines the angle at which fringes appear but also their width. For example, a greater wavelength or smaller slit separation will result in wider fringes, and vice versa. Understanding the relationship between the wavelength and the slit separation is essential for anyone studying wave optics and the nature of light.

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Most popular questions from this chapter

Mourning doves have a small patch of iridescent feathers. The color is produced by a \(330-n m-\) thick layer of keratin \((n=1.56)\) with air on both sides that is found around the edge of the feather barbules. For what wavelength or wavelengths would this structure produce constructive interference?

Pigments don't survive fossilization; even though we have fossil skin from dinosaurs, we don't know what color they were. But fossilization does preserve structure. Specimens from a rare cache of 50 -million-year-old beetle fossils still show the microscopic layers that produced structural colors in the living creatures, and we can deduce the colors from an understanding of thin- film interference. One fossil showed \(80 \mathrm{nm}\) plates of fossilized chitin (modern samples have index of refraction \(n=1.56\) ) embedded in fossilized tissue (for which we can assume \(n=1.33\) ). What is the longest wavelength for which there is constructive interference for reflections from opposite sides of the chitin layers? If this wavelength is enhanced by reflection, what color was the beetle?

A cross section of a scale from the wing of a blue morpho butterfly reveals the source of the butterfly's color. As Figure \(\mathrm{P} 17.69 \mathrm{b}\) shows, the scales are covered with structures that look like small Christmas trees. Light striking the wings reflects from different layers of these structures, and the differing path lengths cause the reflected light to interfere constructively or destructively, depending on the wavelength. For light at normal incidence, blue light experiences constructive interference while other colors undergo destructive interference and cancel. Acetone fills the spaces in the scales with a fluid of index of refraction \(n=1.38 ;\) this changes the conditions for constructive interference and results in a change in color. The change in color when acetone is placed on the wing is due to the difference between the indices of refraction of acetone and air. Consider light of some particular color. In acetone, A. The frequency of the light is less than in air. B. The frequency of the light is greater than in air. C. The wavelength of the light is less than in air. D. The wavelength of the light is greater than in air.

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?

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?
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