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2.71/2.710 Optics
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Practice Problems for the Final Exam Posted Monday, May
Sprin 13, 201
g ’ 3
1. (Pedrotti dist is seen w
ant point hose a
sou
rce
A glass plate is sprayed with of light i
uniform opaque particles. When a
consider Fraunhoffer diffraction through random gratings, and use Babinet’s
ngular width is a
s obs bout
erve 2
d !
lo
. Estimate the size of the particles. (H
oking through the plate, a diffuse h int:
alo
Answer:
principle)
T
circul
he diffra ar apert
ction pa ures
t of t
tern he
of a same size
n opaque circul
!
in a !!
!
n ot
a herw
r part ise opa
icle is complementary to that due to
Under the Fraunhofer condition (
!!!
!
!!
!! (^)!
!
que screen. !!! !!
! !!
exp (−𝑖𝑘(𝜃! 𝑥 + 𝜃!
! !! !
!
!
Where 𝜃! ≈ (^)! , 𝜃! ≈ (^)!
For the given problem, we may further assume E(x, y) is a plane wave at normal incidence, and the transmission function t(x, y) for a single as:
particle can be expressed
𝑥!^ + 𝑦!
Where 𝑅 is the radius of the opaque particles.
𝐸 𝑥!, 𝑦!^ ≈
exp (−𝑖𝑘(𝜃!! 𝑥 + 𝜃!! 𝑦)) 1 − 𝑐𝑖𝑟𝑐(
! ) 𝑑𝑥𝑑𝑦
𝐸 𝑥!, 𝑦!^ ≈
With 𝑥!^ = ! !
! 𝑘^ ,^ 𝑦
𝛿( 𝑘𝑥^2
2 𝜋𝐽 𝑅 𝑘 + 𝑘
2 2
1
2
𝑦 )^ −^ 𝑅^ 2
𝑥 𝑦
𝑅 𝑘𝑥^2 + 𝑘
The ha
𝑦
2
lo is similar to an Airy disc!
We 11_08 provided by Pedrotti:
can evaluate the width of the halo (a second peak) based on the table on Figure
Where 𝛾 = 𝑅 𝑘 2 + 𝑘
From the above table,
𝑥 𝑦
!
Ta find
king the
cent radi
ral us
w of
avel the par
engt ti
h c
at vi le:
sible frequency, λ = 500 nm and given ∆𝜃 = 2 !, we
= 500 𝑛𝑚×
( 2 𝜋)!^
2 =^7463 𝑛𝑚^ =^7.^4 𝜇𝑚
2 = !!^ 𝑅𝜃.
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b) ( point
on
on the
ly ) film
Show at s
distance
that the phase delay of the diverging subject beam, at a
fol the
low film
s w under
hen r<< illumination
. Show also
r that
from the
the amplitude
axis, is given of the
by light
transmitted
𝑙𝑠. This resu by
lts
spherical wavefront, thus a real image on reconstruction.
of the reference beam produces converging
Answer:
derived
The path d in
if a
f rt
erence p (a). Therefore
𝛿 of the divergi the pha
ng beam with respect to the plane wave is se delay is:
!
To a
this system using
nswer last part 𝑘!
of t
he quest 𝑘 !! , 𝑘! =
ion w 𝑘 !! !.
e can calculate the Fresnel diffraction pattern of
𝐸 𝑥′, 𝑦′ ≈ exp ( 𝑖𝑘
! 𝑦 2 𝑧
! ) 1 + cos 𝑘
! exp −𝑖 𝑘𝑥𝑥 + 𝑘𝑦𝑦 𝑑𝑥𝑑𝑦
𝐸 𝑥′, 𝑦′ ≈ ℱ exp ( 𝑖𝑘
𝑥^2
𝑦^2
exp 𝑖𝑘 𝑥 +𝑦 2 𝑠
The Fourier transform of the first term is straight forward:
exp −𝑖𝑘
! .
Likewise, we can express the second and the third term: 1 𝑠 𝑥′!+𝑦′!^1 𝑠 𝑥′!+𝑦′!
th th
e e
opti
rd (^) term indicates a converging wave front towards z=
exp −𝑖𝑘 2 𝑧 𝑠 + 𝑧
exp 𝑖𝑘 2 𝑧 (𝑧 − 𝑠) s ( a real image
) on cal axis.
placed
As shown in the
Figure
a
on surface
B
f
of a
the the
input lens
images L 1 and L
2
The
, 𝑦)and focal
a length
phase mask 𝑡 (𝑥′, 𝑦′) were
2 and a.
th Derive
at of L an
2
2.710 on
is ex ly
pression for t
2 =^. The spacing between L he light distribut
1 and ion on t
L 2 and th
of e
the s
lens
!
n ar
L
e all 2
is f 1 a
he screen.
cree.
b. ( ) Can you suggest a possible application of such an arrangement?
op aq
Figure B. Optical information processing using 2 convex lenses
ue
Answer: a) This problem involves two steps of Fresnel If the illumination is a plane wave, the field b
-‐ diffra ehind t
ction he first
that l
a
wr
ens ca
re casca
itten as:
n b
ded. e
𝐸! 𝑥, 𝑦 = 𝑡!(𝑥, 𝑦)exp −𝑖𝑘
(𝑥!^ + 𝑦!)
We
!
the let it propagate (z=2a)to the front surface of
th
e second lens:
𝐸
ex
𝑥 p
′, (
𝑦 𝑖𝑘
′ 4 𝑎) 4 𝑎 exp^ (^ 𝑖𝑘^
𝑥′^2 +𝑦′ 𝑥𝑥 + 𝑦𝑦 𝑥 +𝑦 4 𝑎
2 ) 𝑒𝑥𝑝(−𝑖𝑘
′ 2 𝑎
′ )exp ( 𝑖𝑘
2 4 𝑎
2
2
) (^)! 2
𝑡 (𝑥, 𝑦)exp −𝑖𝑘
(𝑥!^ + 𝑦!) 4 𝑎 𝑑𝑥𝑑𝑦
𝐸! 𝑥′, 𝑦′ =
exp (𝑖𝑘 4 𝑎) ( 𝑖𝑘
𝑥′ +𝑦′ 4 𝑎
exp 4 𝑎
) 𝑒𝑥𝑝(−𝑖𝑘
𝑥𝑥′^ + 𝑦𝑦 2 𝑎
′ )𝑡!(𝑥, 𝑦)𝑑𝑥𝑑𝑦
exp (𝑖𝑘 4 𝑎) 𝑥 2 +𝑦 2
Then we can repea
𝐸! t t
𝑥 he a
′, 𝑦′ b
= ove st
4 ep t
𝑎 o t
ex he second
p ( 𝑖𝑘
′ 4 𝑎
′
le
) ns
ℱ(
!!
𝑡! f=a
𝑥, ):
𝑦 )
𝐸! 𝑥′, 𝑦′ = 𝑡!(𝑥′, 𝑦′)exp −𝑖𝑘
𝐸− 𝑥′, 𝑦′
Finally at the detector screen:
𝐸 𝑥", 𝑦
= exp (
" 𝑖𝑘 4 𝑎) 4 𝑎 exp ( 𝑖𝑘 𝑥"^2 +𝑦" 4 𝑎
2 ) 𝑒𝑥𝑝(−𝑖𝑘 𝑥"𝑥′^ + 𝑦"𝑦 2 𝑎
′ )exp (− 𝑖𝑘 𝑥′^2 +𝑦 2 4 𝑎
′ )𝑡 2 (𝑥′, 𝑦′)𝐸− 𝑥′, 𝑦′ 𝑑𝑥′𝑑𝑦′
2a 2a
Output Plane
f=2a (^) f=a
t
t 2 (x’, y’)
1 (x, y)
We can rewrite the sine in the transmission function using exponentials:
sin 𝑎𝑥 = 2
𝑒!"#^ −
Equation (1) then becomes:
This informs us that the diffraction pattern consists of three beams with incident
!! !
! (^) !"#$%
2
!!!" !"!^ #$!!^! − 𝑒 !!!" !"!^ #$!!^! (2)
angles 𝜃, 𝜃 + , 𝜃 −. (sin 𝜃 ≈ 𝜃) The int 𝐼! ∝ 𝐸!
ensit
!! !!
𝐸!^!^ 𝑡!𝑒
!
! !
∗ !
y of t he field behind the grating:
! !"#$%^ + !!^ 𝑒!!!"^
!"# !
$! !
! − 𝑒!!!"^
!"#$! !
!!
! !
!!!! !
! !"#$%
!! 𝑒!!!!"^
!"# !
$
!!
!!^! − 𝑒!!!!" !"#!^ !!!^!
B) Using back fo
geometric cal plane, neglecting for
optics, determine the location of the resulting spots on the
of the lens. Use your result to show that the wavelength detection limi
now the diffraction effect due to finite width D t, Δλ, of
Answer:
this spectrometer is inversely proportional to grating period.
Using matrix method, we obtain:
!" ! =!
Therefore, x
!"
!"
!"
out f^ α^ in. for^
!!
!"
=
! !"# 𝑓𝜃,^ 𝑓^ 𝜃^ +^
!
!
!"!
. or
! ,^ 𝑓
!
, ! !.
Whe e is approximately the pixel size of the C
!" =^ ±^ 𝑑𝜆^ =^ ±^! 𝑑𝑥
screen.
r
! CD camera placed on the
C)
dx
width D of the lens, estimate the resulting spot width of the diffracted
( 2.710 only ) Now taking into account of the diffract
λ
ion effect due to finit order a
e
the back focal plane at monochromatic wavelength minimum spectral resolution Δλ of the grating spectrometer.
. Evaluate again the
t
Answer:
approximately:
Under 1D approximation, the diffracted spot at back focal plane is
𝐼± ∝ 𝑡!^ !𝑠𝑖𝑛𝑐
! {𝑥
Or 𝛿𝑥 > 𝜆 ! !
Combined with part b), we obtain: 𝑑𝜆 = ± ! ! 𝑑𝑥^ >^ ±𝜆^
! !
