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normal in the
j
th layer. Thus, for the first boundary (the one between the air and the first
layer)
n
n
i
1
01
=
sin
sin
,
θ
for the second boundary
n
n
2
1
1
2
=
sin
sin
,
and so on. Finally, for the last boundary
n
n
N
N
f
0
=
sin
sin
,
Multiplying these equations, we obtain
n
n
n
n
n
n
n
n
N
iN
f
1
0
2
1
3
2
0
1
1
2
2
3
F
H
G
I
K
J
F
H
G
I
K
J
F
H
G
I
K
J
F
H
G
I
K
J
=
F
H
G
I
K
J
F
H
G
I
K
J
F
H
G
I
K
J
F
H
G
I
K
J
""
sin
sin
sin
sin
sin
sin
sin
sin
.
We see that the L.H.S. of the equation above can be reduced to
n
0
/
n
0
while the R.H.S. is
equal to sin
i
/sin
f
. Equating these two expressions, we find
sin
sin
sin
,
θθ
f
ii
n
n
=
F
H
G
I
K
J
=
0
0
which gives
i
=
f
. So for the two light rays in the problem statement, the angle of the
emerging light rays are both the same as their respective incident angles. Thus,
f
= 0 for
ray
a
,
(b) and
f
= 20° for ray
b
.
(c) In this case, all we need to do is to change the value of
n
0
from 1.0 (for air) to 1.5 (for
glass). This does not change the result above. That is, we still have
f
= 0 for ray
a
,
(d) and
f
= 20° for ray
b
.
Note that the result of this problem is fairly general. It is independent of the number of
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Light

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