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1051-455-20073 Solution Set
- Consider monochromatic light of wavelength λ 0 incident on a single slit of width b along the x-axis
and infinite length along y. The light is observed in the Fraunhofer diffraction region at a distance
L. Derive an approximate expression for the angular width at half-maximum irradiance of the central
peak of the diffraction spot (i.e., the FWHM).
If observed in the Fraunhofer diffraction region, the observed irradiance is proportional to the squared
magnitude of the appropriately scaled Fourier transform:
g [x, y; λ 0 , z 1 ] ∝
z 1
exp
+2πi
μ
z 1
λ 0
− ν 0 t
· F
ξ =
x
λ 0 z 1
, η =
y
λ 0 z 1
The object function is:
f [x, y] = RECT
h
x
b
i
· 1 [y]
Its Fourier transform is:
F [ξ, η] = |b| · SINC [bξ] · δ [η]
= |b| ·
sin [πbξ]
πbξ
· δ [η]
where δ [η] is a 1-D Dirac delta function along the vertical direction, which has finite area and infin-
itesimal support along the vertical direction; we may think of δ [η] as concentrating all of the energy
onto the x-axis. Substitute the variables into this expression:
F
ξ =
x
λ 0
z 1
, η =
y
λ 0
z 1
= |b| ·
sin
h
πb
x
λ 0 z 1
i
πb
x
λ 0 z 1
· δ
y
λ 0
z 1
sin
π
x
λ 0 z 1
b
π
x
λ 0 z 1
´ (^) · δ
y
λ 0
z 1
The irradiance is the square:
I [x, y] ∝ |g [x, y]|
2
∝ SINC
2
x
¡ λ 0 z 1
b
The full width at half maximum is the width of the SINC
2 function between the locations where its
amplitude is +
1
2
SINC
2 [u] =
sin
2 [πu]
(πu)
2
=⇒ SINC [u] =
sin [πu]
πu
= 0.^7071
The half angle θ 1
2
is the value of u that satisfies this relation. We know that SIN C [0] = 1 and
SINC [1] = 0. You can look up the appropriate value of u or solve for it by iteration:
u = 0 .5 =⇒
sin
π
2
π
2
π
= 0.^637
u = 0 .4 =⇒
sin [0. 4 π]
(0. 4 π)
= 0.^757
u = 0 .45 =⇒
sin [0. 45 π]
(0. 45 π)
u = 0 .44 =⇒
sin [0. 44 π]
(0. 44 π)
= 0.^711
u = 0 .445 =⇒
sin [0. 445 π]
(0. 445 π)
u = 0 .4425 =⇒
sin [0. 4425 π]
(0. 4425 π)
= 0.^708
I’ll call that close enough: 0. 4425 π
= 1.^390 radians
SINC
2
x
λ 0 z 1
b
x
λ 0 z 1
b
=⇒ x
λ 0
z 1
b
=⇒ θ 1
2
x
z 1
= 0.^4425
λ 0
b
=⇒ θ = 2 · θ 1 2
λ 0
b
This often is rounded upward so that the full width at half maximum is approximately
θ
λ 0
b
NOTE (again) that the width of the diffraction spot varies as the reciprocal of the slit width.
- Monochromatic light with wavelength λ 0 is incident on a circular aperture of diameter d. The circularly
symmetric diffraction pattern observed at a distance L in the Fraunhofer diffraction region has radius
r from the central maximum to the first zero:
r
= 1.^22
Lλ 0
d
(this distance is the separation required of the diffraction patterns from two point sources for them to
be distinguished under the Rayleigh criterion for resolution)
(a) Compare this result to the linear distance from the central maximum to the first zero for a square
aperture of width equal to the diameter d of the circular aperture.
The radius of the first zero of the diffraction pattern of the circular aperture is
r
Lλ 0
d
The radius to the first zero of a square aperture is the value of x where:
SINC
x
Lλ 0
d
= 0 =⇒ x =
Lλ 0
d
Lλ 0
d
(b) If the monochromatic light at λ 0 illuminates a circular lens of diameter d and focal length f , the
Fraunhofer diffraction pattern is observed at the focal plane so that L
= f and :
r
= 1.^22
f λ 0
d
If observed in blue visible light, find an approximate relation between the diameter of the diffrac-
tion spot and the focal length of the lens. This is a very convenient “rule of thumb” for imaging
systems.
r
f λ 0
d
=⇒ 2 r
f λ 0
d
For blue light, λ 0
= 400 nm =⇒^2.^44 λ 0
=⇒ 2. 44 λ 0
= 1000 nm^
= 1^ μm
2 r
= 1 μm ·
f
d
We often define the focal ratio (f-number, f/#) of the lens as the ratio of the focal length to the
diameter, so the diameter of the diffraction spot measured in micrometers is approximately equal
to the f/# of the lens;
2 r
= f /# [^ μm]
a smaller diameter lens leads to a larger diffraction spot.
- Compare the diameters of the diffraction spots for telescopes with primary optics having diameters
d 1 = 200 in (Hale Telescope on Palomar Mountain) and d 2 = 90 mm (Questar).
2 rPalomar = 1 μm ·
f
200 in ·
1000 mm
39 .37 in
= 1. 97 × 10
− 7 · f
2 r Questar
= 1 μm ·
f
90 mm
= 1.^11 ×^10
− 5 · f
the two optics had the same focal length, the diffraction spot of the Palomar telescope would be smaller
by about a factor of 56.
This is as far as you "had" to go, but you can go farther. The f/numbers of the two telescopes are very
different; the Palomar primary mirror is f/3.3 =⇒ f =3. 3 · d, while the focal length of the Questar
is about 1300 mm =⇒ f /# =
1300
90
= 14, so the diffraction spots have radii approximately equal to
- 3 μm for Palomar and 14 μm for the Questar, which differ by a factor of about 4 instead of 56.
