Chapter 38 IQ Questions - 38.2.1 Effective double slit...

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Unformatted text preview: 38.2.1 Effective double slit model Represent a single-slit of width d by two effective point- sources A and B. Determine the path difference A1’ at the first minimum. Choose one: 1 2 3 4 ‘ Al’ ‘ N4 ‘ 1/2 3N4 l. Hint: At first minimum the phase difference is kA1’21t. Extra: Show that the angle 01’: 1. Id. Explanation: From the hint, A1’2nf(2nil)=l/2. Extra: From the geometry, the angle: 01’=A1’JAB=(L/2)J(d/2)= 1. Id. Digression: We observe that this agrees with the - 2 f intensity expression for the single slit: LB) = M , 1(0”) (Mr where the first minimum is at B=21t=kA1. With A1=L 81=Alfid= ’ . 38.2.3 Single—slit: first minimum \\\\\ \\\\\ \\\\\i\\\\ \ Consider a single slit experiment. Estimate angle 0 at first minimum, for a:0.5mm, 1:500nm, L:10m. Use small angle approx. esza zy/L. Choose one: :06) Mus/2) Hint: m =W—2)2, 13:1;0. Explanation: The first minimum occurs at [322“, or D=1.. So 9 ~ Dxa:500x10'9x(0.5x10'3)=10'3. Ans:3. 38.2.2 IIIo in single—slit experiment ,1 ,1 ,1 ,1 ,1 ,1 ,1 ,1 ,1 ,1 10 ,1 5 ,1 ,1 ,1 ,1 ,1 ,1 ,1 ,1 Consider a single slit experiment. At first minimum the quantities shown in the sketch are: yl, A1 and Blzk A1. Find [5 at y=ylf6. Hint: B=kA. Assume small angle approximation: OzNa :yI’L. fl _ sin2(Bf 2) 1(0”) ' (Biz)2 Explanation: szAzkaylekayIHGL). At the first minimum: B1=2rt=kayllL So B=|3116=Ttl3. Ans=4. Extra: We leave it as an exercise for the reader. Extra: Show that NA1=1/6, & = 9an. 38.2.4 Single—slit: Intensity \\\\\\\\\\ \<\D\\\\\\\\\\\ Given A=?t/3. Estimate 1/10. Choose one: 1 2 3 4 | I/Io ~l ~3/4 ~1/2 ~1/3 1w) sinzwm) Hint: [(00)=W, B=kA. Explanation: For A=N3, [3:271/3. &: sin 2(13 / 2) =(V3/2)2/( n/3)2z3/4. 1(00) (13/2)2 Ans=2. 38.4.1 N slits 1st min. ‘ # of slits angle Phasor diagram ‘ 2 slits ¢1=n 4 slits .3 1:31.12 (114 + Nf16)n 2n/N Explanation: By inspection, - for the 2 slits, 2¢1=2n, (9121:, o for the 4 slits, 4¢I1=2n, $121120, o for the N slits, N¢1=2n, or ¢1=21UN. 38 .4 .3 Double—slits—finite—width pattern \\\ '\\\\\\'g\\\\\\l\‘ §\ For the double—slits-finite-width setup: “d” is slit—distance, “a” slit-width. Incident light has a wavelength 1.. Denote ¢=kA=kd9 and szae. The intensity is given by: 2 ’ B sin lop—’B) = cos2 9 A Here the “double—slit Ill—pattem” 1(0,0) 2 % oscillates within the “single—slit B—pattem”, with the latter serves as an envelope (dotted distribution above). If d=6a, number of zeros within the dotted central eak is 1 2 3 4 “m- Hint: First minimum of single—slit is at B=2n, or 015:21t/ka =Ma, that of doudble-slit is at ¢I=n, or Bldzn/(kd)=L/(2d). Extra: What is the corresponding number, if d=2.5a? Explanation: For d=6a, 01d =lx‘(12a)= 915 /12. For double slits, zeros are at: ¢=[1,3,..., 11]1t., or 6: i [1/12,..., 11/12] 015. There are 12(:2><6) zeros within central peak. Extra: For d=2.5a, 01d=l/(5a). Within the central peak there are 4(:2><2) zeros, i.e. at Old: i[1/5 , 3/5] 615. 38.4.2 Phase diffference between two end rays Consider the phasor diagram for the case of first minimum. For N>2, the first minimum occurs when the contributing vectors form a closed N-sided polygon. For N slits, define |3(=¢Iems) to be the angle measured from E1 Consider the 6 slits case. Sketch the phasor diagram which corresponds the first minimum. The angle [3 for this case is given b : Extra: What is the B for N slits? Explanation: From sketch B=2n—m’3 — Extra: For N slits, |3= 2n—2m’ N. Notice that B for single—slit—finite—width, N—>°°. This leads to |3=2IL 38.6.1 Two polarizer: 5’ 300 #1 #2 O Consider the setup shown. Incident beam with intensity IEl is polarized along the y—axis. Assume #1 transmision axis is 300 with respect to the y—axis, and that of #2 is along the x—axis. Find the final intensity 12. Choose one: Him: Polarized light I=Iocoszoc. Explanation: Il=IDcos2300=SIUf4.12=Ilcos2600=310/16. Ans = 1. 38.6.2 Two polarizen 30D #2 0 Consider the setup shown. Incident beam with intensity In is unpolarized. Assume #1 transmision axis is along the x- axis and #2 is 30'J with respect to y—axis. Choose one: 2 3 4 310/4 [0/2 31014 11/4 311M 31114 Hint: Polarized light I=IDcos20L, Unpolarized light I=Iof2 Explanation: Since the incident light is unpolarized, 11:10/2. When the intermediate ray passes through the second polarizer, 12:11c0826002I1l’4. Ans : 1. ...
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Chapter 38 IQ Questions - 38.2.1 Effective double slit...

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