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Chapter 4: Problem 110

The central diffraction peak of the double-slit interference pattern contains exactly nine fringes. What is the ratio of the slit separation to the slit width?

Short Answer

Expert verified
The ratio of the slit separation to the slit width is given by: \[\frac{d}{a} = \frac{9\lambda}{2\arcsin\left(\frac{\lambda}{a}\right)}\]

Step by step solution

01

Finding the width of the central peak

The width of the central peak is defined as the angular range between the first minimum on either side of the central maximum. This can be found using the equation: \[\theta_{min} = \arcsin\left(\frac{m\lambda}{a}\right)\] where \(m\) is the order of the minimum (in this case, \(m=1\)), \(\lambda\) is the wavelength of the light, and \(a\) is the width of a single slit. To find the angular width of the central peak, we can calculate the difference in angles of the first minimum on either side: \[\Delta\theta = 2\theta_{min} = 2\arcsin\left(\frac{\lambda}{a}\right)\]
02

Finding the angular separation of two fringes

The angular separation between any two adjacent fringes in the interference pattern can be found using the equation: \[\delta\theta = \frac{\lambda}{d}\] where \(d\) is the separation between the two slits.
03

Relating the width of the central peak to the number of fringes

Given that the central diffraction peak contains exactly nine fringes, we can set up a proportion to relate the width of the central peak and the angular separation of two fringes: \[\frac{\Delta\theta}{\delta\theta} = \frac{9}{1}\]
04

Solving the ratio of slit separation to slit width

Substituting the expressions for \(\Delta\theta\) and \(\delta\theta\) into the proportion, we get: \[\frac{2\arcsin\left(\frac{\lambda}{a}\right)}{\frac{\lambda}{d}} = \frac{9}{1}\] Now, we need to solve for the ratio \(\frac{d}{a}\): \[\frac{d}{a} = \frac{9\lambda}{2\arcsin\left(\frac{\lambda}{a}\right)}\] Since we only need the ratio of slit separation to slit width, we can leave the answer in this form. Therefore, the ratio \(\frac{d}{a}\) is given by: \[\frac{d}{a} = \frac{9\lambda}{2\arcsin\left(\frac{\lambda}{a}\right)}\]

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Most popular questions from this chapter

The 305 -m-diameter Arecibo radio telescope pictured in Figure 4.20 detects radio waves with a 4.00 -cm average wavelength. (a) What is the angle between two justresolvable point sources for this telescope? (b) How close together could these point sources be at the 2 million lightyear distance of the Andromeda Galaxy?

How far apart must two objects be on the moon to be resolvable by the 8.1-m-diameter Gemini North telescope at Mauna Kea, Hawaii, if only the diffraction effects of the telescope aperture limit the resolution? Assume \(550 \mathrm{nm}\) for the wavelength of light and \(400,000 \mathrm{km}\) for the distance to the moon.

A double slit produces a diffraction pattern that is a combination of single- and double-slit interference. Find the ratio of the width of the slits to the separation between them, if the first minimum of the single-slit pattern falls on the fifth maximum of the double-slit pattern. (This will greatly reduce the intensity of the fifth maximum.)

If you and a friend are on opposite sides of a hill, you can communicate with walkie-talkies but not with flashlights. Explain.

An amateur astronomer wants to build a telescope with a diffraction limit that will allow him to see if there are people on the moons of Jupiter. (a) What diameter mirror is needed to be able to see \(1.00-\mathrm{m}\) detail on a Jovian moon at a distance of \(7.50 \times 10^{8} \mathrm{km}\) from Earth? The wavelength of light averages \(600 \mathrm{nm}\). (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?
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