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Unformatted text preview: 37.2.1 Double slits:
Qualitative expections Definitions: 1., wavelength of light, L slit—screen
distance, b spacing between adjacent maxima. Statements:
0 a: A bigger L leads to a bigger b.
o b: A bigger 1. leads to a bigger b. o c: A bigger d leads to a bigger b.
Which set of the statements is correct? 1 2 3 4
a&bonl a&conl b&conl a,b,c Hint: Small angle approx. ezA/dzny. Explanation: Maxima occur at the path differences A20,
1., 21., . So the adjacent path difference is at 8A=k,
where the small angle approximation gives SAIdzﬁylL, or b=LL/d. This implies a. is correct, b. is correct and c. is
incorrect. Ans = 1. 37.3.2 Double slits & plastic sheet Consider the double slit experiment, where slit #1 is
covered by a plastic with thickness t and index of
refraction n. O’ is where the phase difference of two rays
shown is zero. Given the extra phase due to the passage of the plastic
¢md=kt(n1 )=4.81t. Determine A.
Choose one:  1! “int: (ltkL2'(kL1+ ¢med):kA' ¢med' Explanation: From the given, ¢=kA—¢med=0. This
allows us to solve for A, which leads to A: (bmed Matt )=
2.41.. Ans=2. 37.3.1 Double slit experiment Consider the setup of a double slit experiment. Find the
secondminimum phase angle :1) and the path difference A. Hint: The intensity I=Iﬂcos (111/2).
Explanation: By inspection, the second minimum
occurs at ¢I=31t, or A=3U2. See sketch below. Ans=3. 2nd min. \ I] 21 41 37.3.3 Double slits: from A to I center Consider the setup of a double slit experiment. Denote the
intensity at the center of the screen to be 10. Then for a path difference A=3U6, the corresponding intensity is:  Hint: The intensity I=Iocos2(¢l!2), ¢=kA, k=2nfL
Extra: What is the least positive value of A which leads
to I=IDJ4? Explanation: The phase angle difference ¢=(2n/l.)(l./6)
=:I't!3. So I=cos2(nf6)10=310f4. Extra: A=U3 leads to ¢=2n/3, which gives If Io=coszol2=cosznf3:1/4. We leave it as an exercise for
the reader to show that this is the minimum positive value
of A which gives the correct intensity. 37.4.1 Double slits with two wavelengths Consider the double slit experiment setup with two
incident waves. A: lA2400nm, and 1st min at y:bA. B: lB=6OOnm, and 1st min at ysz.
In small angle approximation, determine the ratio bB/bA. Hint: First minimum occurs at phase angle difference:
0=kA=Tt, A2112. Extra: Show that the interference pattern under the water
should shrink. Explanation: Bszzb/L. With A2112, bzALx‘d:
lL/(2d). So bB/bAle/1A=600f400=1.5. Extra: Take for instance case A, where L2400nm, and submerge the entire setup under the water. This leads to
the reduction of the wavelength. Being opposite to the
transition in going from case A to case B, the interference
pattern in the water should correspondingly shrink. 37.4.3 Double slits 8: plastic sheet n1,d Consider double slit setup. Lower slit is covered
by a plastic with a thickness d=2p., index of
refraction nl:1.5. Incident wave has wavelength l=0.5p.. Phase angle ¢=¢2—¢1 at 0 is: Hint: ¢= timed =kd(n1—1).
Explanation: rpmd=(2n/l)d(nl—1)=(2tt/0.5) X 2
x (1.51):4n. Ans:4. 37.4.2 Double slit experiment Consider the setup of a double slit experiment. Find the
thirdminimum phase angle I11 and path difference A. Hint: The intensity I=Iﬂcos2(¢/2). Plot [/10 vs ED.
Explanation: By inspection, the minimum sequence is at
¢/2=m’2, 3nf2, Sit/2. So the third minimum occurs at
19:511., or A=(¢/21t)1.=51./2. 37.4.4 Six slits 1 t ' .
l #of slits i Phamdiag i
‘ 2s11ts '31:“ ‘ " i
.._ 4 slits ¢l=nl2 : ﬂ
6 slits ‘31:? Extra: What is the $1 for N slits? Explanation: For number of slits equal to 3 or greater,
the first minimum occurs when the polygon is completed. By inspection this occurs in general at ¢1=2Itf N. For the present case N=6, or ¢1=n/3. Ans = 3. Extra: For N slits, N11) 1227:, or ¢1=2n/ N. 37.5.1 Direct ray & reﬂected ray Consider the superposition of a direct my, #1 and a reﬂected ray, #2 at P, where reﬂection is by a mirror at A. Find kA values which lead to maxima. “int '1’: ¢path + Qreﬂl ' 0reﬂ2 Extra: Find the M values which lead to minima.
Explanation: For the present set up, IIIPath=kA. The phase angle contributed by the reflection is: 0—It=1r..
So maxima occur where the phase angle up: 0, 2n, 47L, . . .=kA+t't. In other words kA= n,3n,51't Extra: The minima occur at kA= 21:, 41c, 67!. 37.6 . 1 Wedge—shaped film 1 2 P Air film is formed by 2 glass plates. Interference pattern is
due to 2 rays as shown. (The sketch is not drawn to scale.
The angle between the two plates should be very small.)
Near the contact point P, should it be close to a maximum
or a minimum? Hint: til: $135131 + Wrem ‘ 'l’relel Explanation: The expression given in the hint leads to
0=(2m"l)(2h) + 0—1t. Near h=0, tum, or the intensity is
close to a minimum. Ans=2. 37.6.2 Counting Dark fringes Given: t=1.6n,7L=0.5n. Determine number of dark fringes in the interval 0A. The dark fringe at O is the
ﬁrst fringe. There aﬁer a dark fringe is included in the count only when the miminum point is included. Hint: Ndark= Integer(¢/2n + 1/2), with ‘1’: ‘l’path + l‘l’reﬂl ' ¢reﬂ2 & “round down “116”: eg
Integer(3.9)=3. Extra: Verify your counting scheme with the consistency
check that, at O, as expected NdadFI.
Explanation: The expression given in the hint leads to
¢/2n= 2U?» +l/2 = 2 x 1.6 /0.5 + 0.5 = 6.9. Or Nda¢=lnteger(6.9+0.5)= 7. Ans=3.
Extra: Let us do the consistent check. At 0, ¢=it. Ndark = Integer (1/2 +1/2) =1. This is expected. 37.6.4 A thin plastic film t,n A light ray with wavelength 1. passes from A to B through
plastic sheet of thickness t and index of refraction n. Find the phase difference Ant: at B, with and without the
film. ' Hint: Phase angles through the film
it =kt, tun =kntznkt. Explanation: The phase difference at B is due to the
phase difference travelling through the thickness of the film, with and without the film, which is given by At: on
— ¢n =(n—1)kt. Ans : 2. 37.6.3 Thin Film Coating l l 111:1.2
t 112:1.6 113:1.5 The incident ray is essentially J. to the surface. Determine re ﬂ. 1 2 3 4 ¢reﬂ 0 713/2 TE 3112/2 Extra: Based on ¢path=2k2t, show that the smallest
thickness for a maximum is at t= MM: nIM/(4nz).
Explanation: For the present case, ¢reﬂ:ﬂ‘0. Extra: The wavelength in vacuum is 7» = nlkl = nzkz. So
Mann/n2. For the present case, ¢reﬂ=TL The maxima
occur at ¢path = ‘n:, 11:, 311, = 2t >< 211/7”. For the smallest thickness, ¢pam= TE, or t = Ml4= HIM/(4m). 37.6.5 Counting Dark Fringes 1 2 O A Consider a very thin wedge shaped air ﬁlm. (Diagram is
not drawn to scale.) Due to the extra path, ray #2 has an
additional phase ¢oath=2. 111:. Determine number of dark fringes in 0A. The first dark ﬁinge is at O. Thereafter a dark ﬁinge is included in the
count only if a minimum is present. l 2 3 N 2 dark. Hint: N dark= Integer(¢/2n + 1/2), with (I) = ¢Dath+ ¢rem  ¢reﬂ2 & “Round down rule”: e.g.
Integer(3 . 9)=3 Explanation: The expression given in the hint leads to
<>l21t= (I)oath +1/2 = 2.ln+1t=3.lit. Or N dark=lnteger(3. 1/2 + 0.5): 2. Ans=2. ...
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This note was uploaded on 02/04/2012 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas.
 Fall '08
 Turner
 Physics

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