62. (a) The condition (in Eq. 3344) required in the critical angle calculation is
θ
3
= 90°.
Thus (with
2
=
c
, which we don’t compute here),
nn
n
11
2 2
33
sin
=
=
leads to
1
=
= sin
–1
n
3
/
n
1
= 54.3°.
(b) Yes. Reducing
leads to a reduction of
2
so that it becomes less than the critical
angle; therefore, there will be some transmission of light into material 3.
(c) We note that the complement of the angle of refraction (in material 2) is the critical
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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

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