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  Remember, when using θ here, you are calculating in radians.

4.  We are again assuming that D » d, where d is the distance between the two slit sources S 1 and S 2, 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)  Rememer the width of the slit needs to be near the magnitude of the wavelength.

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 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

11.  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?

12.  In each of our eyes there is an aperture, the pupil, through which the light enters.  Aperture is a circular opening light enters (as opposed to a slit)  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.

13.  If we look at two point sources then two diffraction patterns will be formed on the retina and these patterns will overlap.  When we say “point sources,” it means any two, separate objects. Could be people, stars, lights, anything that should appear separate)  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

14. Light from the source S 1 enters the eye and is diffracted by the pupil such that the central maximum of the diffraction pattern is formed on the retina at P 1. Similarly, light from S 2 produces a maximum at P 2. 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.

15.  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.

16.  When we pass light through a slit, the diffraction pattern comes up as lighter and darker lines.  When we diffract light through a circular hole, we have a diffraction pattern that resembles a pebble being tossed into a lake, the diffraction pattern is a series of lighter and darker rings.

21.  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

22.  From n = d sin , if n =1, Sin  =  for small angles (in radians)  i.e.  = /d  where d is the width of the slit through which the light from the sources passes.

23.  However, from the diagram of the two sources above this is just the angle that the two sources form at the slit.  Therefore, two point sources will be resolved when the angle that they form at the slit is greater than or equal to /d

24.  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.

25.  It can be shown that = d sin , where d is the aperture of a rectangular objective (similar to a slit)  It can also be shown that  = 1.22 /b for a circular aperture and a small angle, where b is the diameter of the aperture.  The factor of 1.22 arises from a complex mathematical analysis of diffraction from the circular aperture.

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

27.  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.

28.  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.

29.  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.

30.  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.