instructors solution manual for introduction to optics

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instructors solution manual for introduction to optics

Chapter 1 Nature of LightIII, ,, . ft h 6.63 X1-1. a) A =. = 6.63 X 10 3‘p mv (0.05 kg) (20 m/s)h ___________________6-ÍÌ3 X 10~u.l s_____________= 3.

instructors solution manual for introduction to optics .88 X 10“1,1/’ v2»iE [(2-9.11 X IO-31 kg) (10-1.602 X 1(>-»».1)]1/21.2- p = Encrfiy = nhl' = nhc = 100 (6.63X 10~3

50x 10~8m)1-3. The energy of a photon is given by E =hv= h c/AAl A = 770E = (6 63 x= <2-58 * 10‘l,J)1.6ot W-»J = 101 "v1-4. p= fc’/c= ni r2/c= mr = 2. instructors solution manual for introduction to optics

73 X 10~22 kg-m’8, A = Ạ = /ịí = -^-r, == 2.43 x 10 ,2mp E me2 me1-5. E„_o = mc2 = (9.109x 10-:tl kg) (2.998 x 108m/s)2 = (8.187 X 10~H,J)' t

instructors solution manual for introduction to optics

MeVl.lnJli X JU «J1-6. CP--ỰE2 - m2<■1, whereE = Ek +mci = (i +0.511))MeV. So cp=y/1.5112-0.5112 MeV That is. cp= 1.422MeV and p= 1.422 Mev/e.t t he

Chapter 1 Nature of LightIII, ,, . ft h 6.63 X1-1. a) A =. = 6.63 X 10 3‘p mv (0.05 kg) (20 m/s)h ___________________6-ÍÌ3 X 10~u.l s_____________= 3.

instructors solution manual for introduction to optics - v^/e2) *^2 - 11 ~1 — ( — 1/2) t’^/e2) — 1 ] =-^m V2\ ỵ/1 — v2/(~ i LJ-1-9. The total energy of the proton is.E = Ek 4- m, c2 = 2 X 109(1 60 X18 J )

4- (1.67 X10“27 kg) (3.00 X 108m/s)2=4.71 X 10“10 Jự/;2- w2 e4 [(4.71 X1O-*°J)2-(1.67 X IO"27kg)a(3.00 Xio8m)a]”l/2a)p= “c =3.00 X 10» m/sp=1.49x 10“ instructors solution manual for introduction to optics

lskgin/sI,) A =/i/p = (6.63 X 10“31 J-s)/( 1.49 X 10“"’kg m/s) = 4.45 X 10“16 inc) Aph.Is) (3.00 X 10am/s)/(4.71 X 10-I°) =

instructors solution manual for introduction to optics

4.22 X 10-,°m1-10.,, ,____ Energy Energy =__________( lIKHt w III2) ( Hr 1 HI2)____p ““ hv hcfX ((iG3x 1O-:H J)(3.00 X 10hms)/(550x 10“9in)■, ,, m Ey

Chapter 1 Nature of LightIII, ,, . ft h 6.63 X1-1. a) A =. = 6.63 X 10 3‘p mv (0.05 kg) (20 m/s)h ___________________6-ÍÌ3 X 10~u.l s_____________= 3.

instructors solution manual for introduction to optics X 10sm/s _ ...= Ỉ = 770x.0->m =389x 10 ,Li '= A -= 789x 10 llz1-13. The wavelength of the radio waves is A = c/j/ = (3.00 X 10® in s)/(100 X 106 Hz)

= 3 in. The length of the half wave antenna is then XỊ"2 = 1.5 III.1-14. The wavelength is x = c/v = (3.0 X 10bin/s)/(90 X 106 Hz) = 3.33m. The length instructors solution manual for introduction to optics

of each of the rods is then A/4 = 0.83 m.1-15. a) í = Di/c=(90x 103/3.0 X 10s )s = 3.0 X 10“4s. b) D„ = f,t = (310) (3.0 X IO"4 ) in = 0.10in

Chapter 1 Nature of LightIII, ,, . ft h 6.63 X1-1. a) A =. = 6.63 X 10 3‘p mv (0.05 kg) (20 m/s)h ___________________6-ÍÌ3 X 10~u.l s_____________= 3.