Download University of Leicester Department of Physics and Astronomy

University of Leicester
Department of Physics and Astronomy
Lecture Notes
1st Year Optics
Professor R. Willingale
April 7, 2012
Contents
1 Waves, Rays, Image Formation 112-3
2
1.1
Light Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
1.2
Wavefronts - Huygen’s Principle . . . . . . . . . . . . . . . . . . . . . . . .
4
1.3
Reflection and Refraction . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
1.4
Rays - Fermat’s Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
1.5
Image Formation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
1.6
The shape of a lens from first principles . . . . . . . . . . . . . . . . . . . .
7
1.6.1
The properties required of the lens . . . . . . . . . . . . . . . . . .
7
1.6.2
Deviation by a small angled prism . . . . . . . . . . . . . . . . . . .
8
1.6.3
An equation for the lens surfaces . . . . . . . . . . . . . . . . . . .
9
1.6.4
A spherical approximation . . . . . . . . . . . . . . . . . . . . . . . 10
1
1.7
Lens-maker’s equation using Fermat’s Principle . . . . . . . . . . . . . . . 10
2 Interference and Diffraction 112-4
12
2.1
Diffraction through a slit . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.2
Young’s Slits - Two Source Interference . . . . . . . . . . . . . . . . . . . . 12
2.3
Michelson Interferometer . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.4
Coherence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.5
Phasor Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
2.6
Complex Amplitudes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2.7
Diffraction Gratings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2.8
Interference from Thin Films . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.9
Fraunhofer diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
2.10 Diffraction from a Single Slit . . . . . . . . . . . . . . . . . . . . . . . . . . 19
2.11 Diffraction from a Double Slit . . . . . . . . . . . . . . . . . . . . . . . . . 21
2.12 The Limit of Angular Resolution . . . . . . . . . . . . . . . . . . . . . . . 22
1
Waves, Rays, Image Formation 112-3
The material covered is in the following sections of Tipler:
• 33-1 Wave-Particle duality
• 33-2 Light spectrum
• 33-3 Sources of light
• 33-4 The speed of light
• 33-5 Propogation of light
2
• 33-6 Reflection and refraction
• 34-1 Mirrors
• 34-2 Lenses
1.1
Light Waves
Monochromatic light (1 colour) is a travelling harmonic wave.
It is actually an electro-magnetic wave but for the present analysis, amazingly, this is not
important.
If it is travelling along the x-axis in the +ve direction then it has the mathematical form
ψ(x, t) = A sin(kx − ωt + φ)
(1)
where the frequency is ν = ω/(2π) and the wavelength is λ = 2π/k. φ is an arbitary
constant phase angle.
The speed of the wave is
v = ω/k
(2)
This is called the phase velocity because it is the speed at which the peaks (or troughs)
of the wave move.
The phase velocity of a light wave depends on the medium. In a vacuum it is a constant
v = c. We define the refractive index of a medium as
n=
c
v
(3)
As light propogates the frequency remains constant. Therefore if n varies the wavelength
varies. If n is large then k is large and λ is small or if n is small then k is small and λ is
large.
3
1.2
Wavefronts - Huygen’s Principle
Light always fills some volume in space. So ψ above has a value at every point within a
volume. We can write
ψ(r, t) = A sin(k.r − ωt + φ)
(4)
where r is a position vector and k is called the wavevector. |k| = 2π/λ and the wavevector
points in the direction in which the wave is travelling.
The argument of the harmonic sine function is called the phase angle. If this is kept
constant then a 2-d surface is defined. This surface is called a wavefront. The wavevector
k is perpendicular to the wavefronts.
Huygen’s Principle is a geometrical construction that tells us how wavefronts will move
(but not why). It states that:
Every point on a wavefront acts as a source of spherical secondary wavelets such that after
some time (∆t) the primary wavefront lies on the envelope defined by all the secondary
wavelets. The radius of the secondary wavelets will be v∆t.
If the refractive index varies with position we must take small time steps to get the correct
answer.
1.3
Reflection and Refraction
When light hits a plane interface between 2 media then some of the light is reflected and
some is transmitted. The intensity of the reflected and transmitted beams depends on
the materials and the angle of incidence.
We can use Huygen’s Principle to find the reflection and refraction angles. θr = θ1 and
Snell’s Law
n1 sin θ1 = n2 sin θ2
(5)
The refraction (change in direction from the incident to the transmitted beam) arises
because of the change in wavelength of the light.
Note that Huygen’s principle assumes there is no phase change at the boundary.
4
If n2 < n1 what happens if θ1 > θc where sin θc = n2 /n1 ? TIR.
1.4
Rays - Fermat’s Principle
It is often useful to imagine light as consisting of rays. Such rays will always point in the
direction of the wavevector k.
Fermat’s Principle states that the path a light ray takes is such that the time taken to
travel along the path is a minumum or stationary value compared with neighbouring
paths.
But the time taken to travel a distance d is given by
t=
nd
d
=
v
c
(6)
The product ∆ = nd is called the optical path length or simply the optical path. Since c
is constant Fermat’s Principle also states that the optical path is a minimum or stationary
value along the actual path. If n varies with position then we must integrate along the
path to calculate the optical path length.
∆=
Z
P
n(r)dl
(7)
S
We can use this principle to prove the laws of reflection and refraction. See Tipler 33-8.
1.5
Image Formation
A point source produces a set of spherical wavefronts moving away from the point or
alternatively it produces a set of radial diverging rays.
A real image of such a point source is formed by transforming these into a set of spherical
wavefronts moving towards a point or alternatively a set of radial converging rays.
A virtual image of a point source is formed by transforming these into a set of spherical
wavefronts moving away from another point or alternatively a set of radial diverging rays
from another point.
5
In image formation a set of rays from a source point S travel along different paths to
an image point P. By Fermat’s Principle ALL these paths (rays) MUST have the same
optical path length. Or equivalently they contain the same number of wave cycles. If we
follow the passage of a single wavefront from S using Huygen’s Principle then eventually
that wavefront will converge to a single image point P.
Consider a simple imaging system S to P. The direct path SP is the shortest distance
between the object and image. All other paths are longer. A curved mirror can compensate for this by making the peripheral paths shorter by just the correct amount to satisfy
Fermat’s Principle. A convex lens can compensate for this by increasing the optical path
length for rays near the axis.
A virtual image is seen in a plane mirror. In this case the curvature of the wavefronts or
divergence of the rays is unaltered.
A virtual image is seen when looking through a refracting interface, for example the water
surface of a swimming pool. The images seen are at a different position to the objects.
The image of the bottom of the pool appears to be nearer than the actual bottom of the
pool. In this case the curvature of the wavefronts from the object has been changed by
refraction.
Real and virtual images are formed by spherical mirrors and lenses. In these cases the
curvature of the wavefronts is changed or the divergence of the rays is changed.
Paraxial rays are such that they hit the spherical surfaces close to the normal or alternatively they are nearly parallel to the optical axis which is a normal to the surfaces.
If we consider just paraxial rays then for a mirror of radius of curvature R using the law
of reflection we can show that
1
2
1
1
+ 0 =
=
s s
R
f
(8)
where s is the distance of the object from the mirror and s0 is the distance of the image
from the mirror. f is called the focal length.
Similarly for refraction across a single spherical interface of radius R between two media
of refractive indices n1 and n2 using Snell’s Law we can show
1
n2 − n1
1
1
+ 0 =
=
s s
R
f
6
(9)
A thin lens consists of 2 spherical surfaces of radius of curvature Ra and Rb close together.
In this case we can show
1
1
n1
= (n2 − n1 )(
−
)
f
Ra Rb
(10)
This is the lens-makers’ equation.
Lenses and spherical mirrors can convert plane wavefronts into spherical wavefronts and
vice versa. In such a case the object or the image appears at the primary focus a distance
f from the lens or spherical mirror.
Note that plane wavefronts correspond to a source or image at infinity.
1.6
1.6.1
The shape of a lens from first principles
The properties required of the lens
1) Parallel rays near the optical axis are brought to a common focus at a distance f from
the lens (f is the focal length).
For rays incident at distance x from the optical axis the angle of deviation is given by:
tan(α) =
x
≈α
f
providing x small (paraxial rays).
2) Diverging rays from a point source on the optical axis a distance s from the lens are
converted into converging rays which produce an image of the point at a distance s0 the
other side of the lens.
If a diverging ray makes angle γ with the axis and a converging ray angle β with the axis
then the deviation angle is given by:
α=γ+β
If the rays meet at the lens at a distance x from the axis and the angles are small we can
substitute for the angles:
7
α=
x x
+
s s0
We can substitute for α using the focal length f giving:
x
x x
= + 0
f
s s
which reduces to the Gaussian imaging equation:
1
1
1
= + 0
f
s s
The incoming and outgoing rays intersect at a single plane called the Principal Plane of
the lens. This plane is perpendicular to the optical axis.
In order for this equation to hold the deviation angle at a distance x from the axis is given
by (from above):
α=
1.6.2
x
f
Deviation by a small angled prism
If δ is the prism angle then the deviation of a light ray is given by:
α = (n − 1)δ
where n is the refractive index. Providing δ is small this is independent of the incidence
angle. This is easy to prove using Snell’s Law for refraction.
So if we have two nearly parallel surfaces on a dielectric with an angle δ between them
the deviation of a ray passing through the surfaces is given by the above equation.
8
1.6.3
An equation for the lens surfaces
We can combine the results in the last two sections to calculate the equation of the lens
surfaces which will give the desired result.
Assume the lens consists of two identical surfaces back-to-back. If we take the case of a
convex lens the semi-thickness on-axis with be a maximum h0 . The semi-thickness h will
decrease with radius x until we reach a point where h = 0, the largest possible radius for
the lens. What is the equation of h as a function of x which will produce a focal length
f?
If the angle between the front and back surfaces of the lens is δ(x) the gradient of h is
given by:
δ(x) = −2
dh
dx
The deviation angle is therefore:
α = −2(n − 1)
dh
dx
But this is required to be the ratio x/f for imaging:
x
dh
= −2(n − 1)
f
dx
Rearranging we have:
dh
1
= −x
dx
2f (n − 1)
Integrating this equation and imposing the boundary condition that h = h0 for x = 0
gives:
h = h0 −
x2
4(n − 1)f
9
1.6.4
A spherical approximation
We can approximate the parabola above using a spherical surface of radius R. If ∆ =
h0 − h(x) using Pythagoras we have:
(R − ∆)2 + x2 = R2
Neglecting the second order term in ∆ gives:
∆=
x2
2R
But from the previous section:
∆=
x2
4(n − 1)f
So we have that the radius of the spherical surface must be:
R = 2(n − 1)f
or
1
1
1
= (n − 1)( + )
f
R R
which is the lens-maker’s equation for the case where both surfaces are the same radius
and convex. QED.
1.7
Lens-maker’s equation using Fermat’s Principle
We can derive the lens-maker’s equation directly using Fermat’s Principle. Consider
paths from object to image intersecting at principal plane. Since all these are equivalent
(possible) then All paths have the same optical path length. i.e. path is stationary w.r.t.
intersection point on Principal plane.
10
Let maximum thickness of lens be 2h0 as before. Distance from object to lens surface s
and lens surface to image s0 . We implicitly assume 2h0 << s and 2h0 << s0 .
The optical path from object to sphere touching lens and from image to sphere touching
lens is same for all paths. So we need only consider region close to Principal plane.
The optical path has three components; from sphere about object to plane touching centre
of lens, length t; from plane touching otherside of centre of lens to sphere about image,
length t0 ; region of thickness h0 about Principal plane which contains glass of thickness 2h
and refractive index n. All angles w.r.t. the axis are assumed small such that cos θ ≈ 1.
So optical path is:
∆ = ((2h0 + t + t0 ) − 2h) + 2nh
But for spheres:
x2
t≈
2s
t0 ≈
x2
2s0
where x is the distance of the ray from the optical axis at the lens. Therefore we can
substitute for t and t0 :
∆ = 2h0 +
x2 1
1
( + 0 ) + 2(n − 1)h
2 s s
When x = 0 then h = h0 so ∆ = 2nh0 . We can replace the bracket containing s and s0
using the focal length f as before:
x2
= 2(n − 1)(h0 − h)
2f
This is the same parabola derived using Snell’s law above.
11
2
Interference and Diffraction 112-4
The material covered is in the following sections of Tipler:
• 35-1 Phase difference and coherence
• 35-2 Interference in thin films
• 35-3 Two slit interference pattern
• 35-4 Diffraction from a single slit
• 35-6 Fraunhofer and Fresnel diffraction
• 35-7 Diffraction and resolution
• 35-8 Diffraction gratings
2.1
Diffraction through a slit
When light passes through a narrow slit it spreads out or diffracts reaching regions on
the far side of the slit which are inside the classical shadow. This is a consequence of
Huygen’s Principle. A more detailed discussion of this will be given later.
2.2
Young’s Slits - Two Source Interference
If monochromatic light from a distant point source passes through 2 narrow slits separated
by a small distance d then there is a region on the far side where there are 2 sets of
overlaping wavefronts.
Within this region the wave amplitude is the sum of 2 components, E1 = E0 sin(ωt + φ1 )
and E2 = E0 sin(ωt + φ2 ) where φ1 and φ2 depend on the position. The resultant is
E12 = E1 + E2 = 2E0 cos(δ/2) sin(ωt + φ/2)
(11)
where δ = φ1 − φ2 and φ = φ1 + φ2 . This is still a travelling wave of the same frequency
but now the amplitude depends on δ which in turn depends on position.
The intensity of light (the power level detected or the number of photons detected) is
proportional to the square of the amplitude. If there is one slit I0 ∝ E02 . Therefore
12
I12 = 4I0 cos2 (δ/2)
(12)
If we consider a point far from the slits at angle θ wrt the axis then the path difference
between the slits is ∆ = d sin θ and hence the phase difference is given by
δ = d sin θ
2π
= kd sin θ
λ
(13)
If δ = 2mπ where m is an integer then we see a bright fringe. This is called constructive
interference.
If δ = (2m0 +1)π where m0 is an integer then we see a dark fringe. This is called destructive
interference.
2.3
Michelson Interferometer
A light beam can be split using a semi-reflecting interface between two media. These two
beams can be recombined using mirrors.
Now the path difference between the 2 components depends on twice the difference in the
arm lengths 2(l2 − l1 ). The factor of 2 arises because the light travels to and from the
returning mirrors.
Hence the phase difference for light travelling along the axes is given by
δ=
2π
2(l2 − l1 ) = k2(l2 − l1 )
λ
(14)
Hence if we move one of the mirrors we see interference fringes of the same intensity form
as seen in Young’s slits.
2.4
Coherence
It is implicit in the discussion of interference above that the 2 sources are travelling
harmonic waves of the same frequency. Furthermore the waves are in phase at the slits in
Young’s arrangement and in phase at the beam splitter in the Michelson Interferometer.
13
Real light waves can only approximate this ideal. If the light is a single colour then
the range of frequencies is very small so the first condition is reasonably well satisfied.
However light effectively comes in bursts or wave trains which only last a finite time and
the phase difference between successive bursts is random. Therefore if the path difference
is greater than the length of a typical wavetrain the phase between the 2 sources is random
and the 2 sources are said to be incoherent. Incoherent sources will not interfere and will
not produce fringes. Thus the visibility of interference fringes is a measure of how coherent
the light is.
Note that in all interference experiments the sources which are made to interfere ultimately always originate from the same source. Therefore the visibility of interference
fringes in some experiment is a measure of the coherence of that source. It tells us how
monochromatic the source is and how long the wavetrains from the source are.
If there is more than one source or if the source has significant spatial extent then this
can decrease the coherence. For example in Young’s slits experiment if the angular size
of the source is too large when viewed from the slits the interferece fringes will disappear.
In such a case the slits are exposed to overlapping wavefronts from different parts of the
original source and these produce overlapping interference patterns with different lateral
positions on the screen.
2.5
Phasor Diagrams
Phasor diagrams are often used to illustrate how harmonic waves of a single frequency
add together. See Tipler 35-5.
A phasor is a vector with one end at the origin with a length equal to the amplitude of a
harmonic wave. This vector rotates around the origin at angular velocity ω so after time t
it will have rotated ωt radians. If the harmonic wave is cos(ωt + φ) then at time t = 0 the
phasor makes an angle φ radians with the x-axis. Thus the amplitude of the wave at any
time is represented by the projection of the vector onto the x-axis. i.e. the x-component
of the phasor.
It is conventional to draw phasors at t = 0 or alternatively to rotate the axes so that
the phasor always makes an angle φ with the x-axis. We can find the sum of waves of
the SAME frequency by adding their phasors by vector addition. The amplitude of the
resultant phasor is the amplitude of the resultant harmonic wave.
The intensity of the resultant wave is given by the square of the amplitude of the resultant
phasor. This is proportional to the power in the wave (Watts m−2 say) or the number of
photons s−1 m−2 . The amplitude is actually proportional to the transverse electric field
vector and the energy density is proportional to the square of the transvers electric field
14
vector.
2.6
Complex Amplitudes
It is very often convenient to represent a wave by a complex amplitude.
ψ = A(cos(ωt + φ) + i sin(ωt + φ))
This can then be plotted as a point on an Argand Diagram. This will look exactly the
same as a phasor diagram. The amplitude of the actual harmonic wave is given by the
real part of the complex amplitude in just the same way as it is represented by the
x-component of the phasor.
So addition of harmonic waves can be done by complex addition of their complex amplitudes. Such a complex addition is equivalent to phasor addition.
Finally the intensity of the harmonic wave is given by the square of the modulus of
the complex amplitude. We can calculate this by taking the complex amplitude and
multiplying by the complex conjugate.
I = ψψ ∗
This is one of the aspects of complex numbers which is central to their application in
physics.
In the maths course you will find that
A(cos(ωt + φ) + i sin(ωt + φ)) = A exp i(ωt + φ)
This so-called complex exponential representation of harmonic waves is very powerful and
exploited a great deal in physics.
2.7
Diffraction Gratings
What happens if we replace 2 slits (Young’s Slits) with a series of N narrow, parallel,
equally spaced slits?
15
At a position on a distant screen which subtends angle θ with the normal through the
centre of the slits the path difference between adjacent slits is ∆ = d sin θ where d is the
slit spacing so the phase difference is:
δ = kd sin θ
If the amplitude from each slit is S then ignoring any obliquity factor the total complex
amplitude at angle θ is:
A(θ) = S(1 + exp(iδ) + exp(i2δ) + · · · + exp(i(N − 1)δ))
Summing the geometric series gives:
A(θ) = S
1 − exp(iN δ)
1 − exp(iδ)
So the intensity (amplitude squared) at angle θ is:
I = AA∗ = S 2
sin2 (N δ/2)
sin2 (δ/2)
Principle maxima occur at δ/2 = 0, ±π, ±2π, . . .
Minima occur at δ/2 = ±π/N, ±2π/N, . . .
Secondary maxima occur at δ/2 = ±3π/2N, ±5π/2N, . . .
If N is large only the principal maxima are visible. These correspond to the orders of
diffraction.
The mth order is given by
mλ = d sin θ
For values of θ that satisfy this equation a very large peak is seen. Different wavelengths
will appear at different angles so the grating is said to produce a dispersed spectrum. It
splits up the light into the constituent colours.
How wide are these peaks? We have to change the phase difference between adjacent slits
by
16
∆δ =
2π
N
to reach the first minimum (zero) either side of the peak. We also know by differentiating
the expression for δ above that
dδ
= dk cos θ
dθ
Therefore the change in θ required to reach the minimum for the mth order is
∆θ =
λ
N d cos θm
If N is large this angle is very small and hence the peaks are very narrow. If the peaks
are narrow then the spectral resolution of the grating will be high.
In practice gratings are constructed by scoring or etching grooves in a glass substrate. In
modern gratings the grooves can be optimized in shape or blazed to maximize the intensity
that is diffracted into certain orders for a chosen range of wavelengths.
2.8
Interference from Thin Films
Consider a transparent parallel sided film of dielectric thickness d, illuminated by a
monochromatic source at some distance. Plane wavefronts will hit the film at incidence
angle θi .
Some of the light will be reflected and some will be transmitted into the film. The
refracted wavefronts meet the other side of the film and again a fraction is reflected while
the remainder is transmitted. Finally the beam inside the film will meet the front face
again and a fraction is transmitted to form a beam parallel to the original reflected beam.
We want to calculate the condition for interference in the general case when θi is not zero.
So we have 2 beams reflected from the film, 1 from the front surface and 1 from the rear.
The optical path difference between these 2 components is:
∆ = nf (AB + BC) − n1 AD
17
AB = BC = d/ cos θt ,
AD = AC sin θi = AC(nf /ni ) sin θt using Snell’s law
and AC = 2d tan θt so
∆=
2nf d
(1 − sin2 θt ) = 2nf d cos θt
cos θt
If n1 = n2 then 1 beam suffers an internal reflection and the other beam an external
reflection at a n1 : nf interface which introduces a phase difference of π if there is no
absorption. So the phase difference between the beams is:
δ = (2π/λ)2nf d cos θt ± π
Hence for a maximum d cos θt = (2m + 1)λf /4
where m is an integer and λf = λ/nf .
Note that this formula is very similar to that derived for the interference condition in the
Michelson Interferometer.
Actually the light suffers multiple reflections in such a film. We should sum a large number
of reflection terms. If the surfaces of the film are highly reflecting the sum is similar to the
case of the diffraction grating. Two parallel mirrors form a resonant cavity. Such cavities
are used in lasers to increase the gain and/or to tune the light to a particular frequency.
2.9
Fraunhofer diffraction
Fraunhofer diffraction is the diffraction of plane wavefronts through small diffracton angles. It is a limiting case which is mathematically easy to handle and is very important
in the analysis of optical instrumentation.
We can use lenses to produce an experimental arrangement that approximates the conditions for Fraunhofer diffraction quite accurately. This requires a collimator lens and a
telescope lens. Without these lenses the source and screen must both be a large distance
from the diffracting object so that the wavefronts are approximately planar. Thus using
lenses increases the thoughput of light for the system.
Note that diffraction is actually the propogation of light through or around objects. The
so called diffraction pattern is created by interference at a screen or in some light detector.
18
2.10
Diffraction from a Single Slit
How do we calculate the diffraction/interference pattern? We do it in the same way as for
Young’s slits or a diffraction grating, by summing component amplitudes. We divide up
the slit into a large number of small slits and sum the contributions from all these slits.
Because the wavefront is continuous across the slit in the limit we take the sum of an
infinite number of infinitesimal slits. i.e. we perform an integration. If the the wavefronts
incident on the slit are plane and parallel to the slit plane all the components are in phase
at the slit.
Suppose the slit is width a centred on the origin. Then the wavefronts in region −a/2 <
y < +a/2 are transmitted. At an angle θ wrt the axis the path difference wrt the origin
is
∆ = y sin θ
Therefore phase difference is
δ=
2π
y sin θ
λ
The amplitude from an infinitesimal slit at y of width dy is
dE = Ec
dy
a
where Ec is the amplitude at the centre of the viewing screen (on-axis). This implicitly
assumes that the wavelet amplitude from a narrow slit is proportional to the width,
doubling the slit width doubles the amplitude. So the sum of all the infinitesimal slits is
given by the integral
E=
+a/2
Z
−a/2
Ec
sin(ωt + δ)dy
a
We can change the integration variable from y to δ
E=
Z
+β
−β
Ec
sin(ωt + δ)dδ
2β
19
where
β=
πa sin θ
λ
We can evaluate the definite integral by recalling that
sin(A + B) = sin A cos B + cos A sin B
so that
Z +β
Ec Z +β
E=
(
sin ωt cos δdδ +
cos ωt sin δdδ)
2β −β
−β
Because the limits are symmetrical about the origin and sine is an odd function the second
integral is zero. Therefore
E=
Ec
sin ωt sin δ]+β
−β
2β
E = Ec sin ωt
sin β
β
The intensity of the diffraction pattern is given by the time average of the square of the
amplitude.
I = Ic
sin2 β
β2
The function sin β/β is called the sinc function. Note that at β = 0 we must use
L’Hospital’s rule to evaluate the value
sinc(0) =
cos(1)
=1
1
The zeros of the sinc function occur when sin β = 0 excluding the point β = 0 which is
the peak refered to above. Therefore for the minima
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β = π, 2π, 3π, ..., nπ
where n is an integer. These zeros correspond to
sin θ = n
2.11
λ
a
Diffraction from a Double Slit
We are now in a position to calculate the diffraction pattern from a Young’s slits arrangement in which the slits have a finite width a < d where d is the separation of the centres
of the slits.
We split the slits up into an infinite number of infinitesimal slits as above. The integral
becomes
E=
Z +α+β
Ec Z −α+β
sin(ωt + δ)dδ)
sin(ωt + δ)dδ +
(
2β −α−β
+α−β
where β is defined as above and
α=
πd sin θ
λ
We can evaluate the integrals using the same trigonometric substitution as before.
E = 2Ec sin ωt
sin β
cos α
β
The intensity is therefore given by
sin2 β
I = 4Ic 2 cos2 α
β
This is the product of the single slit intensity pattern and the interference fringes expected
from Young’s slits. The factor of 4 arises because now there are 2 slits which doubles the
amplitude on axis and hence quadruples the intensity on axis.
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Note if the width of the slits (a) is very small the first zero of the sinc function sin θ = nλ/a
occurs at a very large angle and we recover the unmodulated cosinusoidal interference
fringes as expected.
2.12
The Limit of Angular Resolution
Consider a Fraunhofer diffraction setup consisting of a collimating lens (to produce plane
wavefronts) and a telescope lens (to focus the diffracted plane wavefronts). We can consider any aperture between the collimator and telescope as defining the size of the aperture
of the telescope.
If the aperture is a single slit of width a then the first zero of the focused spot (in 1-D)
occurs at an angle sin θ = λ/a. As the width of the slit increases so the width of the
diffraction pattern (the focused spot) decreases.
In practice circular apertures are much more common than slits or rectangular apertures.
The diffraction pattern of a circular hole consists of a bright central spot surrounded by
much fainter rings. In between the bright rings are dark rings. The pattern is calculated
in the same way as for the slit above but the aperture is broken into infinitesimal areas
rather than slits and the integral has to be performed in polar coordinates. The resulting
pattern is called the Airy disk.
The angular distance from the centre of the pattern to the first dark ring is given by
∆θ = 1.22
λ
d
where now d is the diameter of the circular aperture (rather than the width of the slit).
So the change from a 1-D slit to a 2-D circular hole just introduces a numerical factor of
1.22.
Again, as the diameter of the aperture (d) increases so the angular size of the central spot
decreases.
Now suppose a second point source is introduced in the focal plane of the collimator. This
will produce a second image in the focal plane of the telescope.
The Rayleigh criterion states that the two images will be resolved if the centre of the Airy
disk of one lies at an angular radius larger than the first zero dark ring of the Airy disk
of the other.
22
∆θR ≈ 1.22
λ
d
Take a look back at the angular width of the diffraction orders of a grating. We found
that
∆θ =
λ
N d cos θm
The factor N d = a is just the total width of the grating so we have
∆θ =
λ 1
a cos θm
This formula has exactly the same form but now we find a numerical factor 1/ cos θm
which usually lies in the range 1 to 1.5 for small orders m.
The angular resolution of an optical system is always given by an equation of the form
∆θ ≈
λ
d
where d is some aperture size. This is true across the whole electromagnetic spectrum,
from radio waves through to gamma rays.
For example, consider the angular resolution of a satellite dish used to receive radio
transmissions from a TV satellite. If the carrier frequency is 12GHz and the diameter of
the dish is 0.75m what is the angular width of the beam?
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