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Two slits spaced 0.450 mm apart are placed 75.0 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 nm?

Short Answer

Expert verified
The distance is approximately 0.083 mm.

Step by step solution

01

Understand the Concepts

Interference patterns involve alternating light and dark bands. Dark bands (minima) occur at positions where the path difference between light from the two slits is a half-integer multiple of the wavelength, i.e., dsinθ=(m+12)λwhere d is the slit separation, θ is the angle of the dark fringe, m is the order of the minimum (1, 2 for second and third dark lines), and λ is the wavelength of the light.
02

Calculate the Angle for the Second Dark Line

For the second dark line, m=1. Using the formula for the dark fringes, 0.450×103sinθ2=(1+12)(500×109)Solve for θ2:sinθ2=1.5×500×1090.450×1031.67×103Thus, θ2sin1(1.67×103).
03

Calculate the Angle for the Third Dark Line

For the third dark line, m=2. Use the dark fringe pattern formula:0.450×103sinθ3=(2+12)(500×109)Solve for θ3:sinθ3=2.5×500×1090.450×1032.78×103Thus, θ3sin1(2.78×103).
04

Find the Positions on the Screen

Use the small angle approximation, where sinθtanθθ and the formula for distance from the center, y=LtanθLθFor θ2 and θ3 calculated earlier, with L=75.0×102 m:y2=75.0×1021.67×1031.2525×101 mmy3=75.0×1022.78×1032.085×101 mm
05

Determine the Distance Between Second and Third Dark Lines

Subtract the positions of the second and third dark lines:Δy=y3y2=0.20850.125250.08325 mm
06

Conclusion

The distance between the second and third dark lines on the screen is approximately 0.083 mm.

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Key Concepts

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

Young's Double-Slit Experiment
Young's Double-Slit Experiment is a classic physics experiment demonstrating the wave nature of light. When light passes through two closely spaced slits, it creates an interference pattern on a screen placed behind the slits. This pattern consists of alternating bright and dark areas called fringes.
If the light coming through the slits was considered as particles, you would expect two distinct lines on the screen because particles travel straight. However, by observing a pattern of multiple lines, it suggests that light behaves like a wave, experiencing constructive and destructive interference.
Understanding this collaboration of light waves through the slits remains a cornerstone of optical and quantum physics.
Optical Path Difference
The Optical Path Difference (OPD) is crucial in analyzing interference patterns. It refers to the difference in the travel paths of light waves coming from the two slits to a common point on the screen.
As waves travel different distances, the OPD leads to either amplification or cancellation of the waves. This interference is mathematically expressed as
dsinθ=mλfor bright fringes
and
dsinθ=(m+12)λfor dark fringes
Here, d is the slit separation, λ is the wavelength of the light, and m is the fringe order.
The OPD concept helps explain why you see specific patterns of light and dark bands.
Wavelength and Slit Separation
Understanding the relationship between wavelength and slit separation is vital in Young's Double-Slit Experiment. The wavelength λ is the distance over which the wave's shape repeats. In this experiment, knowing the wavelength of the incoming light is essential to calculate where the interference fringes will appear.
Slit separation d is the physical distance between the two slits. This separation plays a critical role in determining the spacing of the interference pattern. As the slit separation increases, the fringes move closer together, and vice versa.
The slit separation and wavelength together shape the entire interference pattern according to the formula:
dsinθ=(m+12)λ
where θ is the angle at which a fringe occurs.
Dark Fringe Calculation
Dark fringes, which are the areas of minima in an interference pattern, occur where the path difference leads to destructive interference. To calculate these dark fringes, the formula dsinθ=(m+12)λ is used.
Here, m denotes the order of the dark fringe being calculated. For example, m=1 for the second dark line and m=2 for the third, and so forth.
Once the angle θ for these dark fringes is found using the above formula, you can convert it to a linear distance y on the screen using y=LtanθLθ.
This conversion is possible due to the small angle approximation where sinθθ, allowing you to find the position where the light waves cancel out, leading to the dark fringes.

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Most popular questions from this chapter

In your research lab, a very thin, flat piece of glass with refractive index 1.40 and uniform thickness covers the opening of a chamber that holds a gas sample. The refractive indexes of the gases on either side of the glass are very close to unity. To determine the thickness of the glass, you shine coherent light of wavelength λ0 in vacuum at normal incidence onto the surface of the glass. When λ0 = 496 nm, constructive interference occurs for light that is reflected at the two surfaces of the glass. You find that the next shorter wavelength in vacuum for which there is constructive interference is 386 nm. (a) Use these measurements to calculate the thickness of the glass. (b) What is the longest wavelength in vacuum for which there is constructive interference for the reflected light?

One round face of a 3.25-m, solid, cylindrical plastic pipe is covered with a thin black coating that completely blocks light. The opposite face is covered with a fluorescent coating that glows when it is struck by light. Two straight, thin, parallel scratches, 0.225 mm apart, are made in the center of the black face. When laser light of wavelength 632.8 nm shines through the slits perpendicular to the black face, you find that the central bright fringe on the opposite face is 5.82 mm wide, measured between the dark fringes that border it on either side. What is the index of refraction of the plastic?

In a two-slit interference pattern, the intensity at the peak of the central maximum is I0 . (a) At a point in the pattern where the phase difference between the waves from the two slits is 60.0, what is the intensity? (b) What is the path difference for 480-nm light from the two slits at a point where the phase difference is 60.0?

A compact disc (CD) is read from the bottom by a semiconductor laser with wavelength 790 nm passing through a plastic substrate of refractive index 1.8. When the beam encounters a pit, part of the beam is reflected from the pit and part from the flat region between the pits, so these two beams interfere with each other (Fig. E35.31). What must the minimum pit depth be so that the part of the beam reflected from a pit cancels the part of the beam reflected from the flat region? (It is this cancellation that allows the player to recognize the beginning and end of a pit.)

Coherent light with wavelength 450 nm falls on a pair of slits. On a screen 1.80 m away, the distance between dark fringes is 3.90 mm. What is the slit separation?

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