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Chapter 32: Problem 9

Sketch roughly the diffraction pattern you would expect for light passing through a square hole a few wavelengths wide.

Short Answer

Expert verified
The diffraction pattern will contain a central bright spot (central maxima), surrounded by symmetric, darker bands (minima). Further away, there will be additional bright spots (secondary maxima), but these will not be as bright as the central region. The size and shape of the pattern will be influenced by the size of the hole and the wavelength of the light.

Step by step solution

01

Identifying the setup for diffraction

The given situation suggest that light is passing through a square hole that is just few wavelengths wide. Hence, begin by visualizing this scenario.
02

Anticipating the general pattern

In the case of light passing through a small square hole, a 2D diffraction pattern is expected. The central region (middle) of the pattern will be the brightest. This is due to the constructive interference happening at this point. This region is typically referred as the central maxima.
03

Proposing the detailed pattern

The pattern appears symmetric along both horizontal and vertical middle lines. This symmetry emerges because of the square shape of the hole. Moreover, around the central maxima, there will be several darker bands (minima) radiating outward. The dark bands appear due to destructive interference.
04

Drawing the pattern

Now, keeping in mind the symmetry and position of the maxima and minima, a sketch can be created. The sketch would show a central bright spot with dark bands radiating outward and symmetric across the central maxima. There may be regions of secondary maxima, which are brighter than most of the other regions, but still darker than the central maxima.
05

Detailing the wavelengths

In the final sketch, the size of the hole and the wavelengths of the light will decide the diffraction pattern details. As the hole is only a few wavelengths wide, the diffraction impact will be highly observable. This means the diffraction pattern will be significantly spread out, with brightly lit, and dark areas (maxima and minima respectively).

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Most popular questions from this chapter

Light is incident on a diffraction grating at angle \(\alpha\) to the normal. Show that the condition for maximum light intensity becomes \(d(\sin \theta \pm \sin \alpha)=m \lambda\).

A double-slit experiment with \(d=0.025 \mathrm{mm}\) and \(L=75 \mathrm{cm}\) uses 550 -nm light. Find the spacing between adjacent bright fringes.

Find the wavelength of light used in a Michelson interferometer if 550 bright fringes go by a fixed point when the mirror moves \(0.150 \mathrm{mm}\).

In deriving the intensity in double-slit interference, why can't you simply add the intensities from the two slits?

A thin-walled glass tube of length \(L\) containing a gas of unknown refractive index is placed in one arm of a Michelson interferometer using light of wavelength \(\lambda\). The tube is then evacuated. During the process, \(m\) bright fringes pass a fixed point in the viewer. Find an expression for the refractive index of the gas.
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