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Topic 11.3 Diffraction.

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1 Topic 11.3 Diffraction

2 Definition Diffraction occurs when a wave encounters a barrier or aperture in its path which causes the wave to bend and spread out. The amount of diffraction is most noticeable when the size of the wavelength is bigger or comparable to the size of the barrier/aperture.

3 Diffraction Pattern around a barrier Note areas of constructive and destructive interference

4 Diffraction around a circular Aperture

5 Diffraction of a Narrow Slit
Consider monochromatic light landing on a narrow slit. On a screen placed far away a pattern of bright and dark regions are noticed. The bright regions (maxima) correspond to areas of constructive interference. The dark regions (minima) correspond to areas of destructive interference.

6 Diffraction at a single slit

7 This intensity pattern arises from the fact that each point on the slit acts, in accordance with Huygen's principle, as a source of secondary wavefronts. It is the interference between these secondary wavefronts that produces the typical diffraction pattern.

8 In particular we consider the light from one edge of the slit to the point P where this point is just one wavelength further from the lower edge of the slit than it is from the upper edge. The secondary wavefront from the lower edge will travel a distance λ/2 further than a secondary wavefront from a point at the centre of the slit. Hence when these wavefronts arrive at P they will be out of phase and will interfere destructively

9 The wavefronts from the next point above the upper edge will similarly interfere destructively with the wavefront from the next point above the centre of the slit. In this way we can pair the sources across the whole width of the slit.

10 if is measured in radians
. From the geometry of the situation it follows directly that we have an intensity minimum when: · sin  = /b or for a small angle  approximately that = /b where b = slit width if is measured in radians

11 We could also have found corresponding results with 2, 3,
We could also have found corresponding results with 2, 3, ... This means that if we let the diffracted rays hit a screen rather far away we will get an intensity maximum at  = 0o and minima to both sides of the central maximum whenever sin  = n/b where nI. The maxima/minima pattern continues but with smaller intensity.

12 Example: A single slit of width 1.50 m is illuminated with light of wavelength 500.nm. Find the angular width of the central maximum. Solution: Remember that the spread of the maximum is from 1st order minima on either side. Since there is a minimum on both sides of maximum,  = 2(19.5o) full angular width = 39.0o

13 Example 2 Light with a wavelength of 580 nm is incident on a slit of width 0.30 mm. The observing screen is placed 2.0 m from the slit. Find the position of the first minima.

14 Resolution

15 So what is it that determines whether or not we see the two stars as a single point source i.e. what determines whether or not two sources can be resolved?

16  Our Eyes In each of our eyes there is an aperture, the pupil, through which the light enters. This light is then focused by the eye lens onto the retina. But we have seen that when light passes through an aperture it is diffracted and so if we look at a point source a diffraction pattern will be formed on the retina.

17 Light from the source S1 enters the eye and is diffracted by the pupil such that the central maximum of the diffraction pattern is formed on the retina at P1. Similarly, light from S2 produces a maximum at P2. If the two central maxima are well separated then there is a fair chance that we will see the two sources as separate sources. If they overlap then we will not be able to distinguish one source from another. From the diagram we see as the sources are moved close to the eye then the angle  increases and so does the separation of the central maxima.

18 Diffraction limits our ability to distinguish two different sources of waves which are observed through the same aperture. After the waves pass through the aperture, each source is represented by a diffraction grating. The patterns may overlap and the two sources cannot be resolved separately.

19 Lord Rayleigh(1842-1919) established a guideline:
Sources can be resolved if they have an angular separation greater than or equal to that between the central peak and the first minimum of each sources diffraction pattern.





24 In diagram 3 the two sources will just be resolved
Since this is when the peak of the central maximum of one diffraction pattern coincides with the first minimum of the other diffraction pattern. This means that the angular separation of the peaks of the two central maxima formed by each source is just the half angular width of one central maximum

25 From m  = b sin , if m =1, Sin  =  (in radians) for small angles
i.e.  = /b where b is the width of the slit through which the light from the sources passes.

26 However, from the diagram of the two sources above this is just the angle that the two sources subtend at the slit. Hence we conclude that two sources will be resolved by a slit if the angle that they subtend at the slit is greater than or equal to /b

27 We started this discussion on optical resolution with the eye as the "slit".
But the eye is not a slit but a circular aperture. So to find the "resolving power" of the eye we have to know the half‑angular width of the central maximum of the diffraction formed by a circular aperture.

28 Calculations For a rectangular slit sin = /b
where b is the aperture width of a rectangular objective

29 For a circular aperture:
Two objects will be resolved if their angular separation is larger than the  given by : In radians:  = 1.22  / b For degrees: sin = 1.22/b (  is the angle subtended at the aperture. )

30 Radians An alternative (and some believe better) way of measuring angles.

31 Example 1 Two light sources are 50. cm apart. They are viewed by the eye at a distance L. The pupil of the eye is 3.0 mm. How large can L be and still have sources seen as separate entities? Use = 500. nm. Using degrees:

32 Example 2 The camera of a spy satellite orbiting at 200 km has a diameter of 35 cm. A) Find angular separation B) What is the smallest distance this camera can resolve on the surface of earth? ( Use  = 500 nm)

33 Example If we take the average wavelength of white light to be 500 nm& the average diameter of the human eye to be 3 mm Using  = 1.22  / b the resolving power of the eye is about 2 x 10‑4 rad.

34 Actually because of the structure of the retina and optical defects the resolving power of the average eye is about 3 x 10‑4 rad. This means that the car is more likely to be 5 km away before your resolve the headlights.

35 Microscopes For a microscope, the actual distance when two point objects are just barely resolvable is known as the resolving power. It is given by R.P. = 1.22  D Where D is the diameter of the aperture As a general rule we can say that it is impossible to resolve details of objects smaller than a wavelength of the radiation being used.

36 Diffraction Pattern Vs. Slit Width

37 Diffraction Pattern Versus Number of Slits

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