Topic 11Topic 11

11.4: Resolution11.4: Resolution

Double SlitDouble Slit

In our analysis of the double slit In our analysis of the double slit interference in interference in WavesWaves we assumed that we assumed that both slits act as point sources.both slits act as point sources.

From the previous figure we see that From the previous figure we see that the nth fringe is formed at point P at an the nth fringe is formed at point P at an angle angle θθ given by given by

θθ = y/D = y/D

We are again assuming that D » d, We are again assuming that D » d, where d is the distance between the two where d is the distance between the two slit sources Sslit sources S11 and S and S22, so that we are , so that we are

able to make the small angle able to make the small angle approximation.approximation.

The condition for there to be a bright The condition for there to be a bright fringe at P is thatfringe at P is that

d sin d sin θθ = n = n λλ oror θθ = n = n λλ/d/d (This is for small angles)(This is for small angles)

Under these conditions we obtain fringes of equal Under these conditions we obtain fringes of equal separation and equal intensity.separation and equal intensity.

A sketch of the intensity distribution of the pattern A sketch of the intensity distribution of the pattern is shown below.is shown below.

However, with our knowledge of single However, with our knowledge of single slit diffraction, we must now consider slit diffraction, we must now consider what effect the finite width of the slits what effect the finite width of the slits has on the interference pattern.has on the interference pattern.

If you set up a demonstration of double If you set up a demonstration of double slit interference then you will notice that slit interference then you will notice that you do not actually get fringes of equal you do not actually get fringes of equal intensity and equal spacing.intensity and equal spacing.

If you cover one of the slits then you If you cover one of the slits then you obtain the single slit diffraction pattern.obtain the single slit diffraction pattern.

The double slit pattern is in fact the sum of The double slit pattern is in fact the sum of the single slit diffraction pattern and the the single slit diffraction pattern and the interference pattern formed by two point interference pattern formed by two point sources.sources.

The next figure shows the overall pattern The next figure shows the overall pattern produced when a distance equal to three produced when a distance equal to three times the slit width separates the two slits.times the slit width separates the two slits.

The interference pattern is modulated by The interference pattern is modulated by the diffraction pattern obtained from one the diffraction pattern obtained from one of the slits.of the slits.

If we keep the slit width constant but If we keep the slit width constant but increase their separation then the increase their separation then the overall diffraction envelope stays the overall diffraction envelope stays the same.same.

However, the interference fringes However, the interference fringes become more closely spaced and the become more closely spaced and the overall effect is therefore to pack more overall effect is therefore to pack more fringes into the central maxima.fringes into the central maxima.

If we make the slits narrower then we If we make the slits narrower then we broaden out the central diffraction broaden out the central diffraction maximum and again more interference maximum and again more interference fringes will be "enclosed" by the central fringes will be "enclosed" by the central maxima.maxima.

To show double slit interference more To show double slit interference more clearly we should therefore use very clearly we should therefore use very narrow slits and separate them by a narrow slits and separate them by a reasonable distance,reasonable distance,

in practice a separation equal to about 14 in practice a separation equal to about 14 slit widths.slit widths.

In this way the central diffraction maximum In this way the central diffraction maximum is very broad and fringes close to the centre is very broad and fringes close to the centre will be nearly of the same intensity.will be nearly of the same intensity.

ResolutionResolution

The astronomers tell us that many of the stars that The astronomers tell us that many of the stars that we observe with the naked eye are in fact binary we observe with the naked eye are in fact binary starsstars

That is, what we see as a single star actually That is, what we see as a single star actually consists of two stars in orbit about a common consists of two stars in orbit about a common centrecentre

Furthermore the astronomers tell us that if we use Furthermore the astronomers tell us that if we use a "good" telescope then we will actually see the a "good" telescope then we will actually see the two starstwo stars

we will resolve the single point source into its two we will resolve the single point source into its two component partscomponent parts

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

  Our EyesOur Eyes

In each of our eyes there is an aperture, the pupil, In each of our eyes there is an aperture, the pupil, through which the light enters.through which the light enters.

This light is then focussed by the eye lens onto the This light is then focussed by the eye lens onto the retina.retina.

But we have seen that when light passes through But we have seen that when light passes through an aperture it is diffracted an aperture it is diffracted

and so if we look at a point source a diffraction and so if we look at a point source a diffraction pattern will be formed on the retina.pattern will be formed on the retina.

If we look at two point sources then two diffraction If we look at two point sources then two diffraction patterns will be formed on the retina and these patterns will be formed on the retina and these patterns will overlap. patterns will overlap.

The width of our pupil and the wavelength of the The width of our pupil and the wavelength of the light emitted by the sources will determine the light emitted by the sources will determine the amount that they overlap.amount that they overlap.

But the degree of overlap will also depend on the But the degree of overlap will also depend on the angular separation of the two point sources.angular separation of the two point sources.

We can see this from the next diagramWe can see this from the next diagram

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.

THE RAYLEIGH CRITERIONTHE RAYLEIGH CRITERION

Rayleigh suggested by how much they Rayleigh suggested by how much they should be separated in order for the two should be separated in order for the two sources to be just resolved.sources to be just resolved.

  If the central maximum of one diffraction If the central maximum of one diffraction pattern coincides with the first minima of the pattern coincides with the first minima of the other diffraction pattern then the two other diffraction pattern then the two sources will just be resolved.sources will just be resolved.

This is known as the Rayleigh Criterion.This is known as the Rayleigh Criterion.

In diagram 3 the two sources will just be In diagram 3 the two sources will just be resolvedresolved

Since this is when the peak of the Since this is when the peak of the central maximum of one diffraction central maximum of one diffraction pattern coincides with the first minimum pattern coincides with the first minimum of the other diffraction pattern.of the other diffraction pattern.

This means that the angular separation This means that the angular separation of the peaks of the two central maxima of the peaks of the two central maxima formed by each source is just the half formed by each source is just the half angular width of one central maximumangular width of one central maximum

From m From m = d sin = d sin , if m =1, Sin , if m =1, Sin = = for for small anglessmall angles

i.e. i.e. = = /d /d where d is the width of the slit through where d is the width of the slit through

which the light from the sources passes.which the light from the sources passes.

  

However, from the diagram of the two However, from the diagram of the two sources above this is just the angle that sources above this is just the angle that the two sources subtend at the slit. the two sources subtend at the slit. Hence we conclude that two sources Hence we conclude that two sources will be resolved by a slit ifwill be resolved by a slit if

the angle that they subtend at the slit is the angle that they subtend at the slit is greater than or equal to greater than or equal to /d/d

We started this discussion on optical We started this discussion on optical resolution with the eye as the "slit". resolution with the eye as the "slit".

But the eye is not a slit but a circular But the eye is not a slit but a circular aperture.aperture.

So to find the "resolving power" of the eye So to find the "resolving power" of the eye we have to know the half‑angular width of we have to know the half‑angular width of the central maximum of the diffraction the central maximum of the diffraction formed by a circular aperture.formed by a circular aperture.

Tricky CalculationsTricky Calculations

It can be shown that It can be shown that = a sin = a sin Where a is the aperture of a rectangular Where a is the aperture of a rectangular

objectiveobjective It can also be shown that It can also be shown that = 1.22 = 1.22 /b for a /b for a

circular apperture and a small anglecircular apperture and a small angle Where b is the diameter of the apertureWhere b is the diameter of the aperture

A circular aperture will resolve two A circular aperture will resolve two sources if the angle that they subtend at sources if the angle that they subtend at the aperture is greater than or equal tothe aperture is greater than or equal to

= 1.22 = 1.22 / b / b As mentioned above the angle As mentioned above the angle is is

sometimes called the sometimes called the resolving powerresolving power but should more accurately be called but should more accurately be called the the minimum angle of resolution.minimum angle of resolution.

Clearly the smaller Clearly the smaller the the greater greater the the resolving power.resolving power.

ExampleExample

If we take If we take the average wavelength of white light to be the average wavelength of white light to be

500 nm500 nm & the average diameter of the human eye to & the average diameter of the human eye to

be 3 mmbe 3 mm Using Using = 1.22 = 1.22 / b / b the the resolving powerresolving power of the eye is about 2 x 10of the eye is about 2 x 10‑4‑4 rad. rad.

So suppose that you are looking at car So suppose that you are looking at car headlights on a dark night and the car is a headlights on a dark night and the car is a distance D away.distance D away.

If the separation of the headlight is say 1.5 mIf the separation of the headlight is say 1.5 m Then the headlights will subtend an angle 1.5/D Then the headlights will subtend an angle 1.5/D

at your eye. at your eye. Your eye will resolve the headlights into two Your eye will resolve the headlights into two

separate sources if this angle equals 2 x 10separate sources if this angle equals 2 x 10‑4‑4 rad. rad. i.e. 1.5/D = 2 x 10i.e. 1.5/D = 2 x 10‑4‑4 This gives D = 7.5 km. This gives D = 7.5 km.

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

Actually because of the structure of the Actually because of the structure of the retina and optical defectsretina and optical defects

the resolving power of the average eye is the resolving power of the average eye is about 3 x 10about 3 x 10‑4‑4 rad. rad.

This means that the car is more likely to be This means that the car is more likely to be 5 km away before your resolve the 5 km away before your resolve the headlights.headlights.

MicroscopesMicroscopes

For a microscope, the actual distance when For a microscope, the actual distance when two point objects are just barely resolvable is two point objects are just barely resolvable is known as the known as the resolving powerresolving power..

It is given byIt is given by R.P. = R.P. = 1.22 1.22 DD

Where D is the diameter of the apertureWhere D is the diameter of the aperture As a general rule we can say that it is As a general rule we can say that it is

impossible to resolve details of objects smaller impossible to resolve details of objects smaller than a wavelength of the radiation being used.than a wavelength of the radiation being used.

Multiple Slit DiffractionMultiple Slit Diffraction

If we add further slits to a double slit diffraction If we add further slits to a double slit diffraction grating at the same slit separation, the principal grating at the same slit separation, the principal maxima maintain the same seperationmaxima maintain the same seperation

As nothing has been changed in the formulaAs nothing has been changed in the formula However the maximas become sharper as will be However the maximas become sharper as will be

shown nextshown next Subsidiary maximas appear between the principal Subsidiary maximas appear between the principal

maximas ( n-2 between each, where n is the maximas ( n-2 between each, where n is the number of slits)number of slits)

If we examine the interference pattern If we examine the interference pattern produced when monochromatic light passes produced when monochromatic light passes through a different number of slits we notice through a different number of slits we notice that as the number of slits increases the that as the number of slits increases the number of observed fringes decreases,number of observed fringes decreases,

the spacing between them increasesthe spacing between them increases and the individual fringes become much and the individual fringes become much

sharper.sharper.

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

The slits are very small so that they can The slits are very small so that they can be considered to act as point sources.be considered to act as point sources.

They are also very close together such They are also very close together such that d is small (that d is small (≈≈ 10 10-6-6m).m).

Each slit becomes a source of circular Each slit becomes a source of circular wave fronts and the waves from each wave fronts and the waves from each slit will interfere.slit will interfere.

Let us consider the light that leaves the slit at an Let us consider the light that leaves the slit at an angle angle θθ as shown. as shown.

The path difference between wave 1 and wave 2 The path difference between wave 1 and wave 2 is d sinis d sinθθ and if this is equal to an integral number and if this is equal to an integral number of wavelengths then the two waves will interfere of wavelengths then the two waves will interfere constructively in this direction.constructively in this direction.

Similarly wave 2 will interfere constructively with Similarly wave 2 will interfere constructively with wave 3 at this angle, and wave 3 with 4 etc., wave 3 at this angle, and wave 3 with 4 etc., across the whole grating.across the whole grating.

Hence if we look at the light through a Hence if we look at the light through a telescope, that is bring it to a focus, then telescope, that is bring it to a focus, then when the telescope makes an angle when the telescope makes an angle θθ to the to the grating a bright fringe will be observed.grating a bright fringe will be observed.

The condition for observing a bright fringe is The condition for observing a bright fringe is therefore d sin therefore d sin θθ = n = nλλ

Suppose we use light of wavelength 500 nm Suppose we use light of wavelength 500 nm and suppose that d = 1.6 x 10and suppose that d = 1.6 x 10-6-6m.m.

Obviously we will see a bright fringe in the Obviously we will see a bright fringe in the straight on position straight on position θθ = 0 (the zero order).= 0 (the zero order).

ExampleExample

The next position will be when n = 1(the first order) The next position will be when n = 1(the first order) and substitution in the above equation gives and substitution in the above equation gives θθ = = 18°.18°.

The next position will be when n = 2 (the second The next position will be when n = 2 (the second order) and this give order) and this give θθ = 39°. = 39°.

The next position will be when n = The next position will be when n = 33 (the (the thirdthird order) and this give order) and this give θθ = = 7070°.°.

For n = For n = 44, sin θ is greater than 1 so with this set up , sin θ is greater than 1 so with this set up we only obtain we only obtain 77 fringes, one zero order and fringes, one zero order and threethree either side of the zero order.either side of the zero order.

The calculation shows that the separation of the The calculation shows that the separation of the orders is relatively large.orders is relatively large.

At any angles other than 18° or 38° At any angles other than 18° or 38° or 70 or 70 °° the the light leaving the slits interferes destructively.light leaving the slits interferes destructively.

We can see that the fringes will be sharp since if We can see that the fringes will be sharp since if we move just a small angle away from 18° the light we move just a small angle away from 18° the light from the slits will interfere destructively.from the slits will interfere destructively.

We would see 7 fringes, (zero, plus 2 from n= 1, 2, We would see 7 fringes, (zero, plus 2 from n= 1, 2, and 3)and 3)

An array of narrow slits such as described above An array of narrow slits such as described above is usually made by cutting narrow transparent lines is usually made by cutting narrow transparent lines very close together into the emulsion on a very close together into the emulsion on a photographic plate.photographic plate.

(Typically 200 lines per millimetre). Such an (Typically 200 lines per millimetre). Such an arrangement is called a diffraction grating. arrangement is called a diffraction grating.

The diffraction grating is of great use in examining The diffraction grating is of great use in examining the spectral characteristics of light sources.the spectral characteristics of light sources.

Spectra and WavelengthSpectra and Wavelength All elements have their own characteristic All elements have their own characteristic

spectrum.spectrum. An element can be made to emit light either An element can be made to emit light either

by heating it until it is incandescent or by by heating it until it is incandescent or by causing an electric discharge through it when causing an electric discharge through it when it is in a gaseous state.it is in a gaseous state.

If for example, the element that you are If for example, the element that you are looking has three distinct wavelengths then looking has three distinct wavelengths then each wavelength will be diffracted by a each wavelength will be diffracted by a different amount in accordance with the different amount in accordance with the equation d sin equation d sin θθ = n = nλλ..

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

White LightWhite Light

If white light is shone through a grating then If white light is shone through a grating then the central image will be whitethe central image will be white

but for the other orders each will be spread but for the other orders each will be spread out into a continuous spectrum composed of out into a continuous spectrum composed of an infinite number of adjacent images of the an infinite number of adjacent images of the slit formed by the wavelength of the different slit formed by the wavelength of the different wavelengths present in the white light.wavelengths present in the white light.

Deriving the FormulaDeriving the Formula

This diagram shows just two slits in the This diagram shows just two slits in the diffraction grating, a distance d apart:diffraction grating, a distance d apart:

The path difference between the two The path difference between the two adjacent paths in direction 1 is one adjacent paths in direction 1 is one wavelength, wavelength, λλ..

The path difference between the two The path difference between the two adjacent paths in direction 2 is 2 adjacent paths in direction 2 is 2 wavelengths, 2 wavelengths, 2 λλ..

The angle between the first-order maximum The angle between the first-order maximum 1 and the zero-order maximum 0 is labelled 1 and the zero-order maximum 0 is labelled θθ11 because it is the first-order maximum. because it is the first-order maximum.

For the second-order maximum 2 the For the second-order maximum 2 the corresponding angle is labelled corresponding angle is labelled θθ22

The next diagrams show the triangles made The next diagrams show the triangles made by the dotted lines, the path differences (by the dotted lines, the path differences (λλ) ) and the distance d between the slits:and the distance d between the slits:

The angles at the top of the triangles are The angles at the top of the triangles are also also θθ11 and and θθ22..

Because this triangle is right-angled, you Because this triangle is right-angled, you can write:can write:

sin sin θθ11 = opposite / hypotenuse = opposite / hypotenuse

λλ/d /d or or λλ = d sin = d sin θθ11 for the first order, for the first order,

and 2 and 2 λλ = d sin = d sin θθ22 for the second order. for the second order.

This can be done for a third-order maximum, This can be done for a third-order maximum, fourth-order and so on.fourth-order and so on.

In fact, one equation can be used for all of In fact, one equation can be used for all of themthem

n n λλ = d sin = d sin θθnn for the nfor the nthth order maximum, order maximum,

where n = 1, 2, 3,where n = 1, 2, 3,