Physics 4510 Optics Homework Assignment Problem Set #5 (Turn in before class by Dec 8th, no late homework will be accepted)

1. (15%) Prove that the secondary maxima of a single-slit diffraction pattern occur

when ββ tan= .

2. (15%) A single-slit Fraunhofer diffraction pattern is formed using white light. For what wavelength does the second minimum coincide with the third minimum produced by a wavelength of 400=λ nm.

3. (20%) For an apodized slit which has a transmission function of

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−=

2

2exp21)(ay

ayf

π,

proof that the Fraunhofer diffraction pattern is

)8

exp()(22akdyeyfE iky −=∝ ∫

∞

∞−

by using the relation π=∫∞

∞−

− due u2

.

Note that the diffraction pattern has no sidebands.

4. (15%) Consider a double slits aperture, each slit has a width b, and are separated by a distance h. Prove that the intensity of the Fraunhofer diffraction pattern is

γββ 22

0 cos)sin(II =

where θβ sin21 kb= and θγ sin

21 kh= .

5. (15%) Continue on problem #4, the first term of the diffraction pattern of a

double slits aperture 2)sin(ββ typically accounts for diffraction pattern of a single

slit, and the terms corresponds to the interference between the two slits. Missing orders occur at those values of

γ2cosθsin which satisfy, at the same time, the

condition for interference maxima and the condition for diffraction minima. Show

that this leads to the condition integer.)( =bh

6. (20%) Plot the diffraction pattern )(θI of the double slits aperture for θ from -

180o to 180o in Problem #5, where 10 =I , mb µ1= , mh µ3= and the optical wavelength nm632=λ .

Fo. A

ytLwn,

+' 1;J

l-lw # ss;"61n slit D;{(^ct,"^

S i"r

rL&d F

P*'f) 0

Sf o.'t,;,Fro^"k*(rn

t sin (? 1zt (1 r

t na f , i n r * r .a , :e t # =o

= r. a(f f)#r i f l ir l Dl -

- F s 'n / t( - |

f B c"E[7L t {

I- o

+o

>s*n f(

ze;^ f

*'f )II

T

f le*

= S ; n f

= * . o Ft 4

?=

iirritI t l

i i tI t t

t l lJ t l

iiiI t l

tititit : l

i l !i i ti t it : i

I t tiltr J i, i ; i

l a t* :i f l

liit*li t r

t ' ii i il l i

iirr : i: i ii i i

! t ii : i:. ,,: i 1

f, Fo" sht d,:{Nno*ta4, tL^ ;^-f*nsr1V

tt(

lin c-p

r , si i{ l ) 'r o t f

)

occLr o-Sl*y f= vv ,T I I r n -_ t l , t Z . - -

* t I s i ' r o

+ b s;no

(t -- * Lsr,, s =

.)\

s i n t r = r n A

+1*r'$r*

s i^6lu

T

p r i n i m a

At min irrra ,

anol,

3 \

+ \ '

b

bb

^ \

s r ' n& = eAs r ' n F = J \ /

e A r

* \

6 un y wt I

l r

ut'urz' A =Voonyrt

To, Fr^^ntrr{^r /,ff^drLn,I

I**4,*> I '*!^.^rf ^ f lE -a ( *=) _1 .1 .L

) ^ a J T e e v o l \

(^ -(+)"- tk? ,I *g

" o l 7I l a

t _Iq J T T

sincr (ry)'*ik.* = (*f+iK%+ (f)' f#t=tt * ' f t1 '+*"

-, ) tr (* ̂ -(ry *i$t-(+e)d a J n . } _ : ,

\ - - ' - _ , , t : , _ ' d r r |

= *H - t r

[ e- (#nsf ^ ,= \ F Q. u

I_e-*T - "f ' , , l f ?-.f ]

r:olot,, ̂f e - X t

J - e d , x = f n- k=o'/ g

E-\-)

l)',U tl*

eE

Tl* F nn^n hrfi* /tfi*.t;t^, p^t*c,t,. is

E e( ( ;F t5{no . ! , ,r=

J J"^L[ u]it e \r ag

: f t

n,kt,'^oort + Jl-t e ,k efls;nu uW

J o

= I f ib^ i [e ls in$i K " ^ o L i " e

o o ( ( t Q s i ^ o )

+ f h+b ,

i K6s,,^ $ tL;kg ,ing) ]

) ^ (

I - ihLq,ur$ 'K(h+bls i*$ i f ihsra$

L K r , " o L e - l + e a J| / ^ i k L s i , i $ i i < h s , . , $ \

r k - ^ e ( e l ) ( \ + e )

4t I KW: )s;"$ iK[hl'k;-'S

e*+t

:b I {nLs,'r$

* ktn^e e e

* rus,ns = 1ll F i v

E o{. fb e' e-

tl.+* iuo*a*#&

i k(*)s'' pq +- ikt*)s;,'s

e

i f khrr"s ht+ kbsns) cor{{rh*^$

* n n l

sin (i

F

i t.l', 5in I = |"

CrS X

t *i"tr

1l*afuru ,

'T -TJ- ".Lo

Fo. Tra^^hr1* d;.ll^r1,"^I'aLl< Ett , +tl-

' irj,,^s;tq

T ( o ) = T o + * o r ' $

r ' o l * r o P = * r b s i n 8 a n * t

M ,ss; 1 o.lrnr oao&fs r^rlrnt(

i nfnn {o't

n11u- yv\^Ki6ry\^

dntg

is

*o

sin F

l , Y t : O , t l , t 7 - -

n = t l , t Z

f , = i r l

r

57

{t\

*h

,t ifi.".*un h a l n t n t , + tvl

,^ \rr r \

YN= - =

n

ru\^x Ic*= f I

V -- rnnm i r \ f + )\ F "

= nTT

in*tq,q.r-a h

or

+ ?rP

hot^ TD

% Problem #6% plotting the double slit diffraction pattern

clear all;

theta_angle = -180:.01:180;theta = theta_angle/180*pi;

lambda = 632e-9;b=1e-6;h=3e-6;

k=2*pi/lambda

beta = .5*k*b*sin(theta);gamma = .5*k*h*sin(theta);

I=(sin(beta)./beta.*cos(gamma)).^2;

plot(theta_angle, I);xlabel('Angle');ylabel('Diffraction Pattern');

-200 -150 -100 -50 0 50 100 150 2000

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Angle

Diff

ract

ion

Pat

tern