Topic 11

1. Topic 11 • 11.4: Resolution

2. Double Slit

3. In our analysis of the double slit interference in Waves we assumed that both slits act as point sources. • From the previous figure we see that the nth fringe is formed at point P at an angle θ given by • θ = y/D

4. We are again assuming that D » d, where d is the distance between the two slit sources S1 and S2, so that we are able to make the small angle approximation. • The condition for there to be a bright fringe at P is that • d sin θ = n λ or θ = n λ/d • (This is for small angles)

5. Under these conditions we obtain fringes of equal separation and equal intensity. • A sketch of the intensity distribution of the pattern is shown below.

6. However, with our knowledge of single slit diffraction, we must now consider what effect the finite width of the slits has on the interference pattern. • If you set up a demonstration of double slit interference then you will notice that you do not actually get fringes of equal intensity and equal spacing. • If you cover one of the slits then you obtain the single slit diffraction pattern.

8. The double slit pattern is in fact the sum of the single slit diffraction pattern and the interference pattern formed by two point sources. • The next figure shows the overall pattern produced when a distance equal to three times the slit width separates the two slits.

10. The interference pattern is modulated by the diffraction pattern obtained from one of the slits. • If we keep the slit width constant but increase their separation then the overall diffraction envelope stays the same. • However, the interference fringes become more closely spaced and the overall effect is therefore to pack more fringes into the central maxima.

11. If we make the slits narrower then we broaden out the central diffraction maximum and again more interference fringes will be "enclosed" by the central maxima. • To show double slit interference more clearly we should therefore use very narrow slits and separate them by a reasonable distance,

12. in practice a separation equal to about 14 slit widths. • In this way the central diffraction maximum is very broad and fringes close to the centre will be nearly of the same intensity.

13. Resolution • The astronomers tell us that many of the stars that we observe with the naked eye are in fact binary stars • That is, what we see as a single star actually consists of two stars in orbit about a common centre • Furthermore the astronomers tell us that if we use a "good" telescope then we will actually see the two stars • we will resolve the single point source into its two component parts

14. 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?

15.  Our Eyes • In each of our eyes there is an aperture, the pupil, through which the light enters. • This light is then focussed 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.

16. If we look at two point sources then two diffraction patterns will be formed on the retina and these patterns will overlap. • The width of our pupil and the wavelength of the light emitted by the sources will determine the amount that they overlap. • But the degree of overlap will also depend on the angular separation of the two point sources. • We can see this from the next diagram

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. THE RAYLEIGH CRITERION • Rayleigh suggested by how much they should be separated in order for the two sources to be just resolved. • If the central maximum of one diffraction pattern coincides with the first minima of the other diffraction pattern then the two sources will just be resolved. • This is known as the Rayleigh Criterion.

23. 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

24. From m  = d sin , if m =1, Sin  =  for small angles • i.e.  = /d • where d is the width of the slit through which the light from the sources passes.

25. 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 /d

26. 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.

27. Tricky Calculations • It can be shown that  = a sin  • Where a is the aperture of a rectangular objective • It can also be shown that  = 1.22 /b for a circular apperture and a small angle • Where b is the diameter of the aperture

28. A circular aperture will resolve two sources if the angle that they subtend at the aperture is greater than or equal to •  = 1.22  / b • As mentioned above the angle  is sometimes called the resolving powerbut should more accurately be called the minimum angle of resolution. • Clearly the smaller  the greater the resolving power.

29. 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.

30. So suppose that you are looking at car headlights on a dark night and the car is a distance D away. • If the separation of the headlight is say 1.5 m • Then the headlights will subtend an angle 1.5/D at your eye. • Your eye will resolve the headlights into two separate sources if this angle equals 2 x 10‑4 rad. • i.e. 1.5/D = 2 x 10‑4 • This gives D = 7.5 km.

31. In other words if the car is approaching you on a straight road then you will be able to distinguish the two headlights as separate sources when the car is 7.5 km away from you.

32. 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.

33. 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.

34. Multiple Slit Diffraction • If we add further slits to a double slit diffraction grating at the same slit separation, the principal maxima maintain the same seperation • As nothing has been changed in the formula • However the maximas become sharper as will be shown next • Subsidiary maximas appear between the principal maximas ( n-2 between each, where n is the number of slits)

35. If we examine the interference pattern produced when monochromatic light passes through a different number of slits we notice that as the number of slits increases the number of observed fringes decreases, • the spacing between them increases • and the individual fringes become much sharper.

36. We can get some idea of how this comes about by looking at the way light behaves when a parallel beam passes through a large number of slits. The diagram for this is shown in the next slide.

38. The slits are very small so that they can be considered to act as point sources. • They are also very close together such that d is small (≈ 10-6m). • Each slit becomes a source of circular wave fronts and the waves from each slit will interfere.

39. Let us consider the light that leaves the slit at an angle θ as shown. • The path difference between wave 1 and wave 2 is d sinθ and if this is equal to an integral number of wavelengths then the two waves will interfere constructively in this direction. • Similarly wave 2 will interfere constructively with wave 3 at this angle, and wave 3 with 4 etc., across the whole grating.

40. Hence if we look at the light through a telescope, that is bring it to a focus, then when the telescope makes an angle θ to the grating a bright fringe will be observed. • The condition for observing a bright fringe is therefore d sin θ = nλ

41. Example • Suppose we use light of wavelength 500 nm and suppose that d = 1.6 x 10-6m. • Obviously we will see a bright fringe in the straight on position θ= 0 (the zero order).

42. The next position will be when n = 1(the first order) and substitution in the above equation gives θ= 18°. • The next position will be when n = 2 (the second order) and this give θ = 39°. • The next position will be when n = 3 (the third order) and this give θ = 70°. • For n = 4, sin θ is greater than 1 so with this set up we only obtain 7 fringes, one zero order and three either side of the zero order.

43. The calculation shows that the separation of the orders is relatively large. • At any angles other than 18° or 38° or 70 °the light leaving the slits interferes destructively. • We can see that the fringes will be sharp since if we move just a small angle away from 18° the light from the slits will interfere destructively. • We would see 7 fringes, (zero, plus 2 from n= 1, 2, and 3)

44. An array of narrow slits such as described above is usually made by cutting narrow transparent lines very close together into the emulsion on a photographic plate. • (Typically 200 lines per millimetre). Such an arrangement is called a diffraction grating. • The diffraction grating is of great use in examining the spectral characteristics of light sources.

45. Spectra and Wavelength • All elements have their own characteristic spectrum. • An element can be made to emit light either by heating it until it is incandescent or by causing an electric discharge through it when it is in a gaseous state. • If for example, the element that you are looking has three distinct wavelengths then each wavelength will be diffracted by a different amount in accordance with the equation d sin θ = nλ.

46. Also if you shine laser light through a grating you will see just how sharp and spaced out are the maxima

47. White Light • If white light is shone through a grating then the central image will be white • but for the other orders each will be spread out into a continuous spectrum composed of an infinite number of adjacent images of the slit formed by the wavelength of the different wavelengths present in the white light.

48. Deriving the Formula • This diagram shows just two slits in the diffraction grating, a distance d apart: • The path difference between the two adjacent paths in direction 1 is one wavelength, λ. • The path difference between the two adjacent paths in direction 2 is 2 wavelengths, 2 λ.

49. The angle between the first-order maximum 1 and the zero-order maximum 0 is labelled θ1 because it is the first-order maximum. • For the second-order maximum 2 the corresponding angle is labelled θ2