mm
11
2 2
λ
λ
=
for which the smallest possible choices are
m
1
= 5 and
m
2
= 4. Returning to Eq. 3625,
then, we find
()
4
1
10 mm)
sin
sin
sin
0.36
21 .
0.0056 mm
m
d
θ
−
−−
−
⎛⎞
λ
5(4.0×
==
=
=
°
⎜⎟
⎝⎠
(c) There are no refraction angles greater than 90°, so we can solve for “
m
max
” (realizing
it might not be an integer):
max
4
22
sin90
0.0056 mm
11
10 mm
dd
m
−
°
=
≈
λλ
5
.
0
×
where we have rounded down. There are no values of
m
(for light of wavelength
This is the end of the preview. Sign up
to
access the rest of the document.
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

Click to edit the document details