Type your question here An important diffraction pattern in many situations is d

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An important diffraction pattern in many situations is diffraction from a circular aperture. A circular aperture is relatively easy to make: all that you need is a pin and something opaque to poke the pin through. The figure shows a typical pattern.(Figure 1) It consists of a bright central disk, called the Airy disk, surrounded by concentric rings of dark and light.

While the mathematics required to derive the equations for circular-aperture diffraction is quite complex, the derived equations are relatively easy to use. One set of equations gives the angular radii of the dark rings, while the other gives the angular radii of the light rings. The equations are the following:

Diffraction due to a circular aperture is important in astronomy. Since a telescope has a circular aperture of finite size, stars are not imaged as points, but rather as diffraction patterns. Two distinct points are said to be just resolved (i.e., have the smallest separation for which you can confidently tell that there are two points instead of just one) when the center of one point's diffraction pattern is found in the first dark ring of the other point's diffraction pattern. This is calledRayleigh's criterion for resolvability.

Learning Goal: To use the formulas for the locations of the dark bands and understand Rayleigh's criterion of resolvability.

An important diffraction pattern in many situations is diffraction from a circular aperture. A circular aperture is relatively easy to make: all that you need is a pin and something opaque to poke the pin through. The figure shows a typical pattern.(Figure 1) It consists of a bright central disk, called the Airy disk, surrounded by concentric rings of dark and light.

While the mathematics required to derive the equations for circular-aperture diffraction is quite complex, the derived equations are relatively easy to use. One set of equations gives the angular radii of the dark rings, while the other gives the angular radii of the light rings. The equations are the following:

dark rings: sin%u03B8=1.22%u03BB D or 2.23%u03BB D or 3.24%u03BB D ,

bright rings: sin%u03B8=1.63%u03BB D or 2.68%u03BB D or 3.70%u03BB D ,

where %u03BB is the wavelength of light striking the aperture, D is the diameter of the aperture, and %u03B8 is the angle between a line normal to the screen and a line from the center of the aperture to the point of observation. There are more alternating rings farther from the center, but they are so faint that they are not generally of practical interest.

Consider light from a helium-neon laser (%u03BB=632.8 nanometers) striking a pinhole with a diameter of 0.220mm . Part A At what angle %u03B8 1 to the normal would the first dark ring be observed? Express your answer in degrees, to three significant figures. %u03B8 1 = %u2218 SubmitHintsMy AnswersGive UpReview Part Try Again; 5 attempts remaining Part B This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s).

Diffraction due to a circular aperture is important in astronomy. Since a telescope has a circular aperture of finite size, stars are not imaged as points, but rather as diffraction patterns. Two distinct points are said to be just resolved (i.e., have the smallest separation for which you can confidently tell that there are two points instead of just one) when the center of one point's diffraction pattern is found in the first dark ring of the other point's diffraction pattern. This is calledRayleigh's criterion for resolvability.

Consider a telescope with an aperture of diameter 1.02m .

Part D What is the angular radius %u03B8 1 of the first dark ring for a point source being imaged by this telescope? Use 550 nanometers for the wavelength, since this is near the average for visible light. Express your answer in degrees, to three significant figures. %u03B8 1 = %u2218 SubmitHintsMy AnswersGive UpReview Part Part E This question will be shown after you complete previous question(s). Provide FeedbackContinue Figure 1 of 1

Explanation / Answer

Part A)

Apply sin(angle) = 1.22 (wavelength)/D

sin(angle) = 1.22(632.8 X 10^-9)/2.2 X 10^-4

Angle = .201 degrees

Part D)

sin(angle) = 1.22(550 X 10^-9)/(1.02)

angle = 3.77 X 10^-5 degrees

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