The resolving power of an optical
instrument is its ability to separate the images of two objects, which are close together.
Some binary stars in the sky look like one single star when viewed with the naked eye, but
the images of the two stars are clearly resolved when viewed with a telescope.
Why?
The merging of the images in the eye is
caused by diffraction.
If you look at a far-away object, then the image of the object will form a diffraction pattern on your retina. For two far-away objects, separated by a small angle q, the diffraction patterns will overlap. You are able to resolve the two objects as long as the central maxima of the two diffraction patterns do not overlap. The two images are just resolved when one central maximum falls onto the first minimum of the other diffraction pattern. This is known as the Rayleigh criterion. If the two central maxima overlap the two objects look like one.
The width of the central maximum in a diffraction pattern depends on the size of the aperture, (i.e. the size of the slit). The aperture of your eye is your pupil. A telescope has a much larger aperture, and therefore has a greater resolving power. The minimum angular separation of two objects which can just be resolved is given by qmin=1.22l/D, where D is the diameter of the aperture. The factor of 1.22 applies to circular apertures like the pupil of your eye or the apertures in telescopes and cameras.
The closer you are to two objects, the greater is the angular separation between them. Up close, two objects are easily resolved. As your distance from the objects increases, their images become less well resolved and eventually merge into one image.
Problem:
 | A spy satellite travels at a distance of 50km above Earth's surface.
How large must the
lens be so that it can resolve objects with a size of 2mm and thus read a newspaper?
Assume the light has a wavelength of 400nm.
|
When the medium through which a wave travels abruptly changes, the wave may be partially or totally reflected. When studying mechanical waves we found that when a wave pulse traveling along a rope reaches the end of the rope, it is totally reflected. The details of the reflection depend on if the end of the rope is tied down and fixed, or if it is allowed to swing loose.
A wave pulse, which is totally reflected from a rope with a loose end is not inverted upon reflection. The phase shift of the reflected wave with respect to the incident wave is zero. When a periodic wave is totally reflected, then the incident wave and the reflected wave travel in the same medium in opposite directions and interfere.
We observe a similar phenomenon with light waves. When a light wave reflects from a medium with a larger index of refraction, then the phase shift of the reflected wave with respect to the incident wave is p (180o). When a light wave reflects from a medium with a smaller index of refraction, then the phase shift of the reflected wave with respect to the incident wave is zero. The reflected and the incident wave interfere.
Constructive and destructive interference of reflected light waves causes the colorful patterns we often observe in thin films, such as soap bubbles and layers of oil on water. Thin-film interference is the interference of light waves reflecting off the top surface of a film with the waves reflecting from the bottom surface. If the thickness of the film is on the order of the wavelength of light, then colorful patterns can be obtained, as shown in the image below,
Consider the case of a thin film of oil of thickness t floating on water. For simplicity, assume that the light is incident normally, so that the angle of incidence and the angle of reflection are zero
In the air, the light reflecting off the air-oil interface will have a 180° phase shift with respect to the incident light. A 180o phase shift is equivalent to the light having traveled a distance of 1/2 wavelength. In the oil, the light reflecting from the oil-water interface will have no phase shift with respect to the light incident on the interface. For the light reflected off the oil and the light reflected off the water to constructively interfere we need the two reflected waves to have a phase shift of an integer multiple of 2p (360o). If the light reflected off the oil-water interface travels an additional distance equal to 1/2 the wavelength of the light in oil, then the total phase shift with respect to the light reflected off the air-oil interface will be 2p. This happens if the thickness of the film is equal to1/4 the wavelength of the light in oil. We also get constructive interference if the thickness of the film is equal to 3/4, 5/4, etc, the wavelength of the light in oil. For constructive interference we need
2t = (m+1/2)ln, m = 0,1,2,…,
where ln is the wavelength of the light in oil.
In vacuum we have lf = c. In a medium with index of refraction n we have lnf = c/n. The frequency of oscillation is the same in vacuum and in a medium, therefore
ln = l/n.
For constructive interference we therefore need
2noilt = (m+1/2)l, m = 0,1,2,….
Destructive interference occurs when the thickness of the oil film is equal to (1/2)ln, ln, (3/2)ln, etc.
For destructive interference we therefore need
2noilt = ml, m = 1,2,….
If the thickness of the film is(1/4)ln, the phase of the wave reflected off the top surface is shifted by p by the reflection. The phase of the wave traveling through the film is not shifted by reflection off the bottom surface, but the wave travels an extra distance of ln/2. It will therefore be in phase with the wave reflected off the top surface. If, on the other hand, the film thickness is (1/2)ln, then the wave traveling through the film travels an extra distance of 1 wavelength. It will therefore be out of phase with the wave reflected off the top surface and the two waves will cancel each other out.
Waves incident at an angle qi on the air oil interface are refracted as they enter the oil. The angle of refraction qt is found from Snell's law, nairsinqi = noilsinqt. If they are reflected off the second interface, then they travel a distance 2t/cosqt in the oil. When they emerge again from the oil into the air and propagate parallel to the waves reflected at the air-oil interface, then the total optical path length difference is
2noilt/cosqt - 2dtanqtsinqi = 2noilt/cosqt - 2dtanqt(noil/nair)sinqt
= 2noilt/cosqt(1-sin2qt) = 2noiltcosqt.
For constructive interference we therefore need
2noiltcosqt = (m+1/2)l, m = 0,1,2,…,
and for destructive interference we need
2noiltcosqt = ml, m = 1,2,….
Constructive and destructive interference occur at different angles for different wavelength. The observer sees colored bands.
The destructive interference of reflected light waves is utilized to make non-reflective coatings. Such coatings are commonly found on camera lenses and binocular lenses, and often have a bluish tint. The coating is put over glass, and the coating material generally has an index of refraction less than that of glass. Then the phase shift of both reflected waves is 180°, and a film thickness equal to 1/4 of the wavelength of light in the film produces a net shift of 1/2 wavelength, resulting in cancellation. For such non-reflective coatings the minimum film thickness t required is
t = l/4n,
where n is the index of refraction of the coating material. A coating with thickness t = l/4n prevents the reflection of most of the light with a wavelength l close to l = 4nt. The coating does not reflect a specific range of wavelengths. Often that range is chosen to be in the yellow-green region of the spectrum, where the eye is most sensitive. Lenses coated for the yellow-green region reflect in the blue and red regions, giving the surface a familiar purple color.
Problems:
| When sunlight reflects from a thin film of soapy water, the film appears multicolored,
in part because destructive interference removes different wavelengths from the light
reflected at different places, depending on the thickness of the film.
As the film becomes
thinner and thinner, it looks darker and darker in reflected light, appearing black just
before it breaks. The blackness means that destructive interference removes all
wavelengths from the reflected light when the film is very thin. Explain.
| ||
| A thin film of a material is floating on water (n
= 1.33). When the material has a
refractive index of n = 1.20, the film looks bright in reflected light as its thickness
approaches zero. But when the material has a refractive index of n
= 1.45, the film looks
black in reflected light as the thickness approaches zero. Explain these observations in
terms of constructive and destructive interference and the phase changes that occur when
light waves undergo reflection.
|
The colorful patterns that you see when light reflects off a compact disk are produced by thin film interference.
A compact disc is made of a polycarbonate wafer which is coated with a metallic film, usually an aluminum alloy. The aluminum film is then covered by a plastic polycarbonate coating. The coatings are less than 100nm thick and each coating partially reflects and partially transmits incident light. Light rays reflected from different coating boundaries interfere with each other to produce the colorful patterns. The reflectance of the CD is not uniform, because CD disk contains a long string of pits written helically on the disk. These pits encode the information stored on the CD.
Thin film interference is used in industry as a non-contact, non-destructive way to measure film thicknesses.
Link to other Web material:
| Thin film interference |
| Audio compact disks |
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