# Diffraction around an edge and through an aperture

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Diffraction around an edge and through an aperture
Huygens’ principle is consistent with diffraction: The larger the wavelength with respect to the aperture size, the more significant the diffraction

Young’s Double Slit Experiment
If light is a wave, interference effects will be seen, where one part of wavefront can interact with another part. One way to study this is to do a double-slit experiment:

Two-Source Interference Pattern
Nodal points – blue dots Anti-nodal points – red dots

Young’s Double Slit Experiment
If light is a wave, there should be an interference pattern

Path Difference 1) there is no difference 2) half a wavelength 3) one wavelength 4) three wavelengths 5) more than three wavelengths In a double-slit experiment, what path difference have the waves from each slit traveled to give a minimum at the indicated position? Intensity

Path Difference 1) there is no difference 2) half a wavelength 3) one wavelength 4) three wavelengths 5) more than three wavelengths In a double-slit experiment, what path difference have the waves from each slit traveled to give a minimum at the indicated position? 2 3 For Destructive Interference  = 1/2 , 3/2 , 5/2 , 7/2 , … = (m + 1/2)  Intensity /2 3/2 5/2 7/2

Diffraction by a Single Slit or Disk
bright dark Note: There are bright and dark fringes beyond the shadow. These resemble interference fringes of a double slit.

Diffraction by a Single Slit or Disk
Contrary to expectation, an interference pattern is observed because the wave front approaching slit, D, is itself a source of wavelets. It is these individual wavelets that will interfere destructively at the part of the screen where the dark band is observed. Wavelets from the center and edge of the slit Dark Band (Interference of wavelets) Y D D D Path difference = m/2 Wave front going through the slit is made up of many wavelets. Now from the diagram the angle to the first dark band is given by: sinD = (/2) / (D/2) =  / D X If D is small we can use….. similar triangles, for small angles sin D = tan D = Y / X so… Y / X = m  / D or… Y = m  X / D m = 1,2,3……. Therefore for any dark band sin D = m  / D m = 1,2,3…….

Diffraction by a Single Slit or Disk
If we look at the intensity of the bright bands we find that they are not the same. The brightest band is in the center and the bands seem to get dimmer as the angle increases. The diagram shows how the intensity changes with angle. You can see that as sin approaches  / D the intensity falls to zero as expected from our theory. Also note that the second band is much less intense than the central band and is half as wide. Practice Problem Light of wavelength 750 nm passes through a slit 1.0 x 10-3 mm wide. How wide is the central maximum a) in degrees and b) in centimeters on a screen 20 cm away? = 750 x 10-9 m D = 1.0 x 10-6 m X = 0.2 m away m = 1 Width of central maximum is 2 or 2Y a) Width = 2 = 2 sin-1 (m  / D) = 2 sin-1 (1) (750 x 10-9m) / (1.0 x 10-6 m ) = b) Width = 2Y = 2 X tan D = 2 (0.2) tan = m

The diffraction pattern below arises from a single slit. If we would like to sharpen the pattern, i.e., make the central bright spot narrower, what should we do to the slit width? 1) narrow the slit 2) widen the slit 3) enlarge the screen 4) close off the slit

The diffraction pattern below arises from a single slit. If we would like to sharpen the pattern, i.e., make the central bright spot narrower, what should we do to the slit width? 1) narrow the slit 2) widen the slit 3) enlarge the screen 4) close off the slit d q The angle at which one finds the first minimum is: The central bright spot can be narrowed by having a smaller angle. This in turn is accomplished by widening the slit. sin  =  / d

Blue light of wavelength l passes through a single slit of width d and forms a diffraction pattern on a screen. If the blue light is replaced by red light of wavelength 2l, the original diffraction pattern can be reproduced if the slit width is changed to: 1) d/4 2) d/2 3) no change needed 4) 2 d 5) 4 d

Blue light of wavelength l passes through a single slit of width d and forms a diffraction pattern on a screen. If the blue light is replaced by red light of wavelength 2l, the original diffraction pattern can be reproduced if the slit width is changed to: 1) d/4 2) d/2 3) no change needed 4) 2 d 5) 4 d d q d sin = m (minima) If   2 then we must have d  2d for sin  to remain unchanged (and thus give the same diffraction pattern).

When violet light of wavelength 415 nm falls on a single slit,
it creates a central diffraction peak that is 9.20 cm wide on a screen that is 2.55 m away. How wide is the slit? = 415 x 10-9 m Y = 4.6 x 10-2 m X = m away m = 1 D = ? Because X is much bigger than Y, D is small, so we can use….. Y = m  X / D m = 1,2,3……. D = m  X / Y = (1) (415 x 10-9 m) (2.55 m) / (4.66 x 10-2 m) D = x 10-5 m = x 10-2 mm

Rayleigh Criterion for Resolution of Two Diffracted Images
Two images are said to be just resolved when the central maximum of one image falls on the first minimum of the diffraction pattern of the other image. From single slit diffraction sinD =  / D If  << d, sinD ≈ D min =  / D Where min is in radians The apertures of microscopes and telescopes are circular. Analysis shows… min = 1.22  / D

An optical telescope has a 21 cm mirror diameter. To give the same angular resolution as an optical telescope the effective diameter of a radio telescope would have to be…. 1) bigger 2) smaller 3) the same 4) doesn’t matter

An optical telescope has a 21 cm mirror diameter. To give the same angular resolution as an optical telescope the effective diameter of a radio telescope would have to be…. 1) bigger 2) smaller 3) the same 4) doesn’t matter The minimum angle of resolution is given by: Radio waves have a much longer wavelength, so to give the same minimum angle of resolution the dish diameter has to be much larger min =  / d

A scientist wants to observe finer detail on a specimen under a microscope. They could…. 1) Make the objective bigger 2) Use smaller wavelengths 3) Use oil between the specimen and the objective 4) All of the above

A scientist wants to observe finer detail on a specimen under a microscope. They could…. 1) Make the objective bigger 2) Use smaller wavelengths 3) Use oil between the specimen and the objective 4) All of the above The minimum angle of resolution is given by: To make min smaller you would have to make the objective lens bigger or use a smaller wavelength. Oil has a higher n than air so n = air / n. Spherical and chromatic aberration however limit the size of the objective, d. Also UV is absorbed by glass. Electrons can be used because of their wave properties (E = h f) min =  / d

Two of Jupiter’s largest moons (it has 16 or more) are separated
by a distance of 3.0 x 106 km at a time when the planet is 3.3 x 108 km from Earth. a) Would a person be able to resolve both moons with the unaided eye, assuming night-time pupil diameter of 7.5 mm? b) What minimum diameter of mirror would be needed in a radio telescope? Use: light = 550 nm, radio = 21 cm and s = r  = 550 x 10-9 m s = 3.0 x 109 m r = 3.3 x 1011 m d = 7.5 x 10-3 m dmin = ?  = s / r = (3.0 x 109 m / (3.3 x 1011 m) = rad a) min = 1.22  / d = 1.22 (550 x 10-9 m / (7.5 x 10-3 m) = rad Yes  > min , so a person would be able to resolve both moons b) dmin = 1.22  /  = 1.22 (21 x 10-2 m) / = 28 m Now you can see why big dishes are needed for radio telescopes

What is the smallest distance this camera can resolve on the
A camera on a spy satellite orbiting at 200 km has a diameter of 35 cm. What is the smallest distance this camera can resolve on the surface of the earth? Use: light = 550 nm, and s = r  = 550 x 10-9 m s = ? r = 200 x 103 m d = 35 x 10-2 m min = 1.22  / d = 1.22 (550 x 10-9 m / (35 x 10-2 m) = 1.92 x 10-6 rad s = r min = (200 x 103 m)(1.92 x 10-6 ) = m (38 cm) This size of lens would not be good enough to read license plate numbers but good enough to resolve individual people.