Chapter 37 IQ Questions - 37.2.1 Double slits: Qualitative...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5
This is the end of the preview. Sign up to access the rest of the document.

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(n-1 )=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 second-minimum 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 (1.5-1):4n. Ans:4. 37.4.2 Double slit experiment Consider the setup of a double slit experiment. Find the third-minimum 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”: e-g- 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. ...
View Full Document

Page1 / 5

Chapter 37 IQ Questions - 37.2.1 Double slits: Qualitative...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online