Physics 111 Homework Solutions Week #9 - Tuesday Friday, February 25, 2011 Chapter 22 Questions - None Multiple-Choice 223 A 224 C 225 B 226 B 227 B 229 D Problems 227 In this double slit experiment we are given a wavelength of light of 488nm, a slit width for each slit of 08mm a slit spacing of 024mm The angular distance from the central maximum to the 1 st interference maximum is given by The angular distance from the central maximum to the 1 st diffraction minimum is given by Thus in the 1 st diffraction minimum there will be interference fringes to the right to left of the central Including the central maximum we arrive at a total of 7 fringes However the last two lie at the diffraction minimum will not be visible, so we can only see 5 fringes 228 A double-slit experiment with 500nm light In this double slit experiment we are given a wavelength of light of 500nm a slit width for each slit of 01mm a The spacing between the slits is given by: b The angular distance from the central maximum to the 1 st diffraction minimum is given by sin θ = λ/a, the distance is given as
c The distance between bright interference fringes is given as therefore In the 1 st diffraction minimum there are interference fringes on either side of the central maximum for a total of 8 fringes In addition we need to add in the central maximum for a total of 9 fringes However, the two extreme interference maxima are located at the 1 st diffraction minima are not visible, thus we need to subtract 2 The total number of visible interference fringes is 7 229 Double-slit diffraction with a beam of electrons In this double slit experiment we are given beam of electrons incident on a two slits each of width of 01µm separated by 40µm a In order to calculate the de Broglie wavelength for the electrons we need to calculate their momentum hence their velocity Accelerating the electron through a potential difference of 100V does work on the electrons the electrons acquire a speed given by: wavelength is given as Their de Broglie b The width of the central diffraction maximum is twice the distance from the central maximum to the 1 st diffraction minimum on either side, calling it y 1 This distance is given as: Therefore the total width is twice this distance or 48mm c To calculate the number of fringes, we will use the same logic as in problem 8c The distance between adjacent bright interference fringes is given as: The distance from the central maximum to the 1 st diffraction minimum is ½ of the width found in part 9b Thus the number of fringes around the central maximum is fringes plus the central maximum for a total of 9 fringes The two extreme fringes are at a diffraction minimum are not visible Thus there a total of 7 fringes visible
2210 A double-slit experiment with 550nm light a The fringe spacing is given as where for small angles we have For Δm = 1, the spacing is b The width of the diffraction pattern is given by, which is valid for small angles Thus we have for the ½ of the central width twice this value, or 220mm therefore the total width is c The number of fringes about the central maximum is given by Including the central maximum produces 15 total fringes However, the last two are at the diffraction minimum are not visible Thus the number of visible fringes is 13 d In water the wavelength of light changes The wavelength in water is given by Redoing parts a c gives: a The fringe spacing is given as where for small angles we have For Δm = 1, the spacing is b The width of the diffraction pattern is given by, which is valid for small angles Thus we have for the ½ of the central width width is twice this value, or 1652mm therefore the total c The number of fringes about the central maximum is given by Including the central maximum produces 15 total fringes However, the last two are at the diffraction minimum are not visible Thus the number of visible fringes is 13 e If a polarizer is placed in front of each of the slits with their transmission axes at right angles to each other then supposing that the sources are vertically polarized, one of the light beams will be extinguished while the other is transmitted You will see no interference pattern, only a single slit diffraction pattern with a bright
central spot ever decreasing intensities as you move away from the central maximum 2217 A double slit experiment with protons a The wavelength is given by the de Broglie relation We have where the velocity is determined from the potential difference that the protons were accelerated though Accelerating the protons through a potential difference does work on the protons this work changes their kinetic energy hence their speed The speed is given as therefore the wavelength of the proton is b The center-to-center spacing of the constructive interference maxima is, assuming that the small angle approximation is valid, c The width of ½ of the central maximum is given as Therefore the full width is twice this value or 228µm d The number of interference fringes is Including the central maximum produces 9 total fringes However, the last two are at the diffraction minimum are not visible Thus the number of visible fringes is 7 Monday, February 28, 2011 Chapter 24 Questions 241 Based on special relativity we know that as a particle with mass travels near the speed of light its mass increases In order to accelerate this object from rest to a speed near that of light would require an ever increasing force (one that rapidly becomes larger by a factor
of γ) There are no known forces that could accelerate a particle with mass to the speed of light in a finite amount of time with a finite amount of energy So for objects with no rest mass, as they travel at the speed of light, there mass does not increase with increasing speed we avoid these problems of accelerating the massless particles Multiple-Choice - None Problems 241 Relativistic energy momentum for an object of mass m For an object with a m = 1kg rest mass, it has a rest energy of E = mc 2 = 9x10 16 J The Lorentz factor is given by:, with the relativistic momentum p = γmv, the relativistic energy E 2 = p 2 c 2 +m 2 c 4 a For a velocity of 08c, The relativistic momentum energy are therefore, b Following the procedure in part a, for a velocity of 09c,, the relativistic momentum is 619x10 8 kgm/s, the relativistic energy is 207x10 17 J c For a velocity of 095c, γ = 320, the relativistic momentum is 913x10 8 kgm/s, the relativistic energy is 288x10 17 J d For a velocity of 099c, γ = 709, the relativistic momentum is 211x10 9 kgm/s, the relativistic energy is 6387x10 17 J e For a velocity of 0999c, γ = 224, the relativistic momentum is 670x10 9 kgm/s, the relativistic energy is 20x10 18 J 242 For an electron with rest mass 911x10-31 kg, it has a rest energy of E = mc 2 = 8199x10-14 J = 0511 MeV The Lorentz factor is given by:, the relativistic momentum energy are given respectively as p = γmv
a For the electron with a velocity of 08c, γ = 167 The relativistic momentum energy are therefore, Using the fact that 16x10-19 J = 1 ev, the relativistic energy is given as Further it can be shown that the total relativistic energy is given as Thus the relativistic energy can be computed in a more efficient method b Following the method outlined in part a, for a velocity of 09c, γ = 229 The relativistic momentum is therefore, c For a velocity of 095c, γ = 32 The relativistic momentum is therefore, d For a velocity of 099c, γ = 709 The relativistic momentum is therefore, e For a velocity of 0999c, γ = 224 The relativistic momentum is therefore, 244 The relativistic momentum relativistic energy are given as respectively To show the relation between energy momentum, equation (248), we start by squaring the relativistic energy This gives us Next, we use a mathematical trick We add subtract the same quantity from the right h side of the equation we just developed The quantity we want to add subtract is v 2 This produces factoring out a factor of c 2, Exping this result we get more than p 2 c 2 allows us to write Recognizing that the first term is nothing Factoring out a c 2 from the 2 nd term on the right h side give us
The quantity is simply Therefore we arrive at the desired result, 247 For a photon, a its momentum is given by b its wavelength is given by the de Broglie relation c its frequency is given by