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Chapter 4: Problem 1

As the width of the slit producing a single-slit diffraction pattern is reduced, how will the diffraction pattern produced change?

Short Answer

Expert verified
As the width of the slit producing a single-slit diffraction pattern is reduced, the angular position of the minima (dark fringes) increases, resulting in wider spacing between the fringes and a more spread out pattern. The central bright spot will also get broader, and the overall light intensity of the pattern will decrease.

Step by step solution

01

Recall the single-slit diffraction formula

The single-slit diffraction formula relates the slit width, the wavelength of the light, the angular position of the minima (dark fringes), and the order of the dark fringe (m). The formula is written as: \( a \sin{\theta} = m\lambda \) where: - a is the width of the slit - θ (theta) is the angular position of the minima (dark fringes) - m is the order of the dark fringe (m = 1, 2, 3, ...) - λ (lambda) is the wavelength of the light
02

Analyze the relationship between slit width and angular position of the minima

We want to understand how the diffraction pattern changes as the slit width is reduced. By examining the formula, we can see that the slit width (a) and the angular position of the minima (θ) are inversely related. If we decrease the slit width (a), the angular position of the minima (θ) will increase.
03

Understand the meaning of larger angular positions for the minima

A larger angular position for the minima indicates that the dark fringes will be located at larger angles from the center of the diffraction pattern. As a result, the spacing between the fringes will be wider, and the overall pattern will appear to "spread" sideways.
04

Conclude how the diffraction pattern changes with a reduction in slit width

As the width of the slit producing a single-slit diffraction pattern is reduced, the diffraction pattern will change by having wider spacing between the fringes, making the pattern "spread" sideways and appear more extensive. Consequently, the central bright spot will get broader, and the light intensity of the pattern will decrease overall.

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Most popular questions from this chapter

What are the angular positions of the first and second minima in a diffraction pattern produced by a slit of width \(0.20 \mathrm{mm}\) that is illuminated by \(400 \mathrm{nm}\) light? What is the angular width of the central peak?

A microwave of an unknown wavelength is incident on a single slit of width \(6 \mathrm{cm} .\) The angular width of the central peak is found to be \(25^{\circ} .\) Find the wavelength.

(a) Find the angle between the first minima for the two sodium vapor lines, which have wavelengths of 589.1 and \(589.6 \mathrm{nm},\) when they fall upon a single slit of width \(2.00 \mu \mathrm{m} .\) (b) What is the distance between these minima if the diffraction pattern falls on a screen \(1.00 \mathrm{m}\) from the slit? (c) Discuss the ease or difficulty of measuring such a distance.

Crystal lattices can be examined with X-rays but not UV. Why?

An X-ray scattering experiment is performed on a crystal whose atoms form planes separated by \(0.440 \mathrm{nm}\). Using an X-ray source of wavelength \(0.548 \mathrm{nm}\), what is the angle (with respect to the planes in question) at which the experimenter needs to illuminate the crystal in order to observe a first-order maximum?
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