光學作業 應物三甲 0932385 劉俊緯
OPTICS: Problem 7-27
A laser beam from a 1-mW He-Ne laser (632.8 nm) is directed onto a parallel film
with an incident angle of 450 . Assume a beam diameter of 1mm and a film index of
1.414. Determine (a) the amplitude of the E-vector of the incident beam; (b) the angle
of refraction of the laser beam into the film; (c) the magnitude of r’ and t t’, using the
stoke relation and a reflection coefficient , r = 0.280; (d) the independent amplitude of
the first three reflected beam and by comparison with the incident beam, the
percentage of radiant power density reflected in each; (e) the same information as in
(d) for the first two transmitted beams; (f) the minimum thickness of film that would
lead to total cancellation of the reflected beams when they are brought together at a
point by a lens.
1
(a) I   0 c  E0  E0    0 cE02
2
1
1*103W
= *8.85*10 12 *3*10 8 E02
3 2
2
 (0.5*10 )
E0  980
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(b) n1 sin 1  n2 sin 2
1*sin 450  1.414sin 
  300
(c) r = 0.28 = r ' = 0.28
t t’ = 1  r 2
= 0.9216
(d) Er  rE0 = 0.28 * 980 = 274
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Er 2  tt ' r ' E0 = 0.9216*0.28*980 = 253
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Er 3  tt ' r ' E0 = 0.9216* (0.28)3 *980 = 19.8
3
I r1 E 2 r1

 0.078  7.8%
I 0 E 20
I r 2 E 2r 2
 2  0.067  6.7%
I0
E0
I r 3 E 2r 3
 2  0.00041  0.041%
I0
E 0
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(e) Et1  tt ' E0 = 0.9216*980 =903
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Et 2  tt 'r ' E0 = 0.9216* (0.28)2 *980 = 70.8
2
I t1 E 2 t1

 0.849  84.9%
I 0 E 20
I t 2 E 2t 2
 2  0.0052  0.52%
I0
E 0
(f)   2n f t cos t  m
2*1.414 * t * cos 300 = 632.8nm
632.8
t=
= 258nm
6
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