3. Diffraction and Optical resolution


Resolution is defined as the ability of an optical system to differentiate between two closely spaced lines. In terms of the optics, the limit of resolution of a microscope objective refers to its ability to distinguish between two closely spaced Airy disks in the diffraction pattern. The Airy disk represents the absolute limit of resolution achievable by an optical system. (Ultimately, any image is made up of an astronomical number of Airy disks). According to the Rayleigh Criterion, the objects will be resolved when they have a separation of at least R.  In an ideal optical system, the optical resolution is restricted by the numerical aperture (NA) and by the wavelength of light;


Resolution is proportional to the size of the NA and inversely proportional to the wavelength. Therefore, if the highest possible resolution is to be achieved, then either the probing frequency must be increased, or the size of the NA should be increased. Illumination with with ultraviolet light (UV), rather than photopic wavelengths, will permit finer detail, allowing the user to observe better resolution of the subject. However, by decreasing the illumination wavelength, finer and finer detail can be resolved, until a threshold limit is reached, called the diffraction limit. The ideal diffraction limit of an optical microscope is around 200 nm, around half the wavelength of illumination. However, this is an ideal measure which can never be achieved due to imperfections in the optics.

There are two parameters which define the resolution of a digital image, ie. the amount of information that a digital image holds. Firstly, the spatial resolution, which is the number of pixels used to construct an image. Obviously, the more pixels that are present in a given amount of area, the higher the spatial resolution will be. But resolution is also dependent on the dynamic range available to the pixel elements. The dynamic range is the range of intensity values or grey values that the detector can differentiate. The pixels in a digital image will have discreet grey levels that cycle between varying levels from white to black. The cycle rate between black and white levels is called the spatial frequency. When the sampling frequency is reduced, fine details are lost, resulting in aliasing. The greater the dynamic range of a detector, the better the contrast will be. The contrast …..

The resulting image resolution is a measure of the degree to which the digital image represents the fine details of the analogue image recorded by a microscope.

The aspect ratio is the width to height ratio of an image.

3. Diffraction and Optical resolution

Single slit diffraction pattern

Waves being diffracted from a wide slit
Diffraction pattern for a single slit of width larger than the wavelength (d, is greater than, lambda,d>λ).

The diffraction pattern made by waves passing through a slit of width a,a (larger than lambda,λ) can be understood by imagining a series of point sources all in phase along the width of the slit. The waves moving directly forward are all in phase (they have zero path difference), so they form a large central maximum.

Waves passing straight through the slit, all sources in phase.

If the wave travels at an angle theta,θ from the normal to the slit, then there is a path difference x,x between the waves produced at the two ends of the slit.

x, equals, a, sine, theta.  x=asinθ

Waves passing through the slit at an angle with path difference x between top and bottom.

The path difference between the top and middle waves is then

x, prime, equals, start fraction, a, divided by, 2, end fraction, sine, theta.  x′=a/2​sinθ

If the path difference between the top and middle waves is start fraction, lambda, divided by, 2, end fraction,2λ​, then they are exactly out of phase and cancel each other out. This happens to all consecutive pairs of waves (the ones produced by the second source from the top and the second source past the middle etc.) at this angle, so there is no resultant wave at this angle. Thus, a minimum in the diffraction pattern is obtained at

lambda, equals, a, sine, theta.  λ=asinθ

Now the slit can be divided into four equal sections and the pairing of sources to give destructive interference can be repeated for the top two sections, which is identical to the result of pairing off matching sources in the bottom two sections. In the case, we obtain for a minimum (since every pair of waves we consider will destructively interfere due to our choice of geometry and pairing), to givestart fraction, lambda, divided by, 2, end fraction, equals, start fraction, a, divided by, 4, end fraction, sine, theta.


We can then divide the slit aperture into six equal sections, and pair off sources in the top two divisions, then the middle two divisions, and then the bottom two, to give destructive interference for every matched pair. The minima of intensity are obtained at anglesn, lambda, equals, a, sine, theta.  


where n,n is an integer left bracket, 1, comma, 2, comma, point, point, point, right bracket,(1,2,…), but not n, equals, 0,n=0. There is a maximum of intensity in the centre of the pattern. This process only gives the positions of the minima, does not work for positions of the maxima, and so does not give the intensities of the maxima.

These results lead to the diffraction pattern minima shown below, which can be represented as a graph of intensity of the diffracted wave against angle. See a level 6 section for an explanation of this diffraction pattern using phasors.

The figure shows a single slit diffraction pattern, with intensity against distance. The graph is a plot of a sinc(x)^2 function.
Single slit diffraction pattern.
3. Diffraction and Optical resolution


Diffraction is the spreading out of waves as they pass through an aperture or around objects. It occurs when the size of the aperture or obstacle is of the same order of magnitude as the wavelength of the incident wave. For very small aperture sizes, the vast majority of the wave is blocked. For large apertures the wave passes by or through the obstacle without any significant diffraction.

In an aperture with width smaller than the wavelength, the wave transmitted through the aperture spreads all the way round and behaves like a point source of waves (they spread out below).

This is shown in the diagram below:

Wave scattering as it passes through a thin slit.

Figure 1: Single slit diffraction when a wave passes through an aperture with width smaller than the wavelength (d, is less than, lambda,d<λ). For a significant amount of the wave to pass through, the aperture must be close to the size of the wavelength.