ch36 - Chapter 36 1. (a) We use Eq. 36-3 to calculate the...

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1383Chapter 361. (a) We use Eq. 36-3 to calculate the separation between the first (m1= 1) and fifth2(5)mminima:21sin.mDDyDDmmmaaa Solving for the slit width, we obtainaDmmy21640055010510 352 5bgbgchb gmmmmmmmm...(b) Form= 1,sin...ma1 550102 52 21064bgchmmmmThe angle is= sin–1(2.210–4) = 2.210–4rad.2. From Eq. 36-3,almsin1sin30.02.00.3. (a) A plane wave is incident on the lens so it is brought to focus in the focal plane ofthe lens, a distance of 70 cm from the lens.(b) Waves leaving the lens at an angleto the forward direction interfere to produce anintensity minimum ifasin=m, whereais the slit width,is the wavelength, andmisan integer. The distance on the screen from the center of the pattern to the minimum isgiven byy = Dtan, whereDis the distance from the lens to the screen. For theconditions of this problem,sin...ma1 590100 4010147510933bgchmmThis means= 1.47510–3rad andy= (0.70 m) tan(1.47510–3rad) = 1.010–3m.4. (a) Equations 36-3 and 36-12 imply smaller angles for diffraction for smallerwavelengths. This suggests that diffraction effects in general would decrease.
CHAPTER 361384(b) Using Eq. 36-3 withm= 1 and solving for 2(the angular width of the centraldiffraction maximum), we find22sin1la2sin10.50m6.0m9.6.(c) A similar calculation yields 0.23° for= 0.010 m.5. (a) The condition for a minimum in a single-slit diffraction pattern is given byasin=m,whereais the slit width,is the wavelength, andmis an integer. For=aandm= 1,the angleis the same as for=bandm= 2. Thus,a= 2b= 2(350 nm) = 700 nm.(b) Letmabe the integer associated with a minimum in the pattern produced by light withwavelengtha, and letmbbe the integer associated with a minimum in the patternproduced by light with wavelengthb. A minimum in one pattern coincides with aminimum in the other if they occur at the same angle. This meansmaa= mbb. Sincea= 2b, the minima coincide if 2ma= mb. Consequently, every other minimum of thebpattern coincides with a minimum of theapattern. Withma=2, we havemb= 4.(c) Withma=3, we havemb= 6.6. (a)= sin–1(1.80 cm/2.00 m) = 0.516°.(b) For themth diffraction minimum,asin=m. We solve for the slit width:amsin2 441nmsin0.5160.098mm .7. The condition for a minimum of a single-slit diffraction pattern issinamwhereais the slit width,is the wavelength, andmis an integer. The angleismeasured from the forward direction, so for the situation described in the problem, it is0.60° form= 1. Thus,95633 10m6.04 10m .sinsin 0.60ma
13858. Let the first minimum be a distanceyfrom the central axis that is perpendicular to thespeaker. ThensinyDymaa221 2ch(form= 1).Therefore,yDa21Dafvs21100m0.300m2500Hz343m s2151.4m .9. The condition for a minimum of intensity in a single-slit diffraction pattern isasin=m, whereais the slit width,

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Term
Spring
Professor
N/A
Tags
Diffraction, Wavelength, Sin, NM

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