Physics 212
Classical and Modern Physics
Spring 2008
Solutions for HandIn Set #7
A47 – Resolution
I paced off 184 steps to the point where I could just
barely resolve the numerals on the clock, so my dis
tance was
184 paces
×
0
.
75 m
1 pace
+ 30 ft
×
0
.
304 m
1 ft
= 147 m
The diameter of my pupil in bright sunlight is about
0
.
3 cm = 3
×
10

3
m, so for
λ
= 500 nm, my eye’s reso
lution is
θ
= 1
.
22
λ
d
= 2
.
0
×
10

4
rad
.
For the angular size of the separation between numbers
to be the same as my resolution, then
tan (2
.
0
×
10

4
rad) =
x
147 m
,
where
x
is the linear separation between the numerals.
So,
x
= 147 tan (1
.
1
×
10

4
rad) = 3
×
10

2
m
= 3 cm.
A88
In this twoslit problem, we’re looking for the angle be
tween the centeral maximum and the third side maxi
mum.
third side max
central max
θ
At the angle of the third side maximum, Δ
φ
= 6
π
, so
Δ
r
=
Δ
φ
2
π
λ
= 3
λ.
But this path length difference is related to the angle
θ
by
Δ
r
=
d
sin
θ
where
d
is the separation between the slits. So,
sin
θ
=
3
λ
d
=
3(633
×
10

9
)
3
×
10

6
=
0
.
633
and
θ
= 39
.
3
◦
.
A90
We can calculate the path lengths for all three waves
directly from the geometry of the problem:
3 m
3 m
P
5 m
5 m
4 m
The path length difference between the waves from the
top and middle speakers is Δ
r
= 1 m, and we get the
same value for the path length difference between waves
from the middle and bottom speakers.
This gives a
phase difference of
Δ
φ
= 2
π
Δ
r
λ
=
π.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Ladd
 Physics, Light, Orders of magnitude, optical path length, Atot

Click to edit the document details