12. We will make use of arctangents and sines in our solution, even though they can be
“shortcut” somewhat since the angles are small enough to justify the use of the small
angle approximation.
(a) Given
y
/
D
= 15/300 (both expressed here in centimeters), then
θ
=
tan
−
1
(
y
/
D
) = 2.86°.
Use of Eq. 366 (with
a
= 6000 nm and
λ
= 500 nm) leads to
( )
6000nm sin 2.86
sin
1.883rad .
500nm
a
α
π°
π
==
=
λ
Thus,
2
sin
0.256 .
p
m
I
I
⎛⎞
⎜⎟
⎝⎠
(b) Consider Eq. 363 with “continuously variable”
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '08
 Any
 Physics

Click to edit the document details