Kapoor (mk9499) – homework38 – Turner – (60230)
1
This
printout
should
have
14
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
(part 1 of 2) 10.0 points
Light
of
wavelength
580
nm
illuminates
two slits of width 0
.
03 mm and separation
0
.
18 mm.
How many interference maxima fall within
the full width of the central diffraction maxi
mum?
Correct answer: 11.
Explanation:
Let :
λ
= 580 nm
,
d
= 0
.
18 mm
,
and
w
= 0
.
03 mm
.
The angle of the first diffraction minimum
to the width
w
of the slits of the diffraction
grating is
sin
θ
1
=
λ
w
,
and the angle corresponding to the
m
th
inter
ference maxima is
sin
θ
m
=
m
λ
d
.
We require
θ
1
=
θ
m
, so
λ
w
=
m
λ
d
m
=
d
w
.
Therefore the number of fringes in the central
maximum is
N
= 2
m

1
= 2
parenleftbigg
d
w
parenrightbigg

1
= 2
parenleftbigg
0
.
18 mm
0
.
03 mm
parenrightbigg

1
=
11
.
002
(part 2 of 2) 10.0 points
What is the ratio of the intensity of the third
interference maximum to the side of the cen
terline (not counting the center interference
maximum) to the intensity of the center in
terference maximum?
Correct answer: 0
.
405285.
Explanation:
The phase difference between the lights at
centerline and at third interference maximum
is
φ
=
2
π
λ
w
sin
θ
3
=
2
π
λ
w
3
λ
d
=
6
π w
d
,
so the ratio of the intensities is
I
3
I
0
=
sin
φ
2
φ
2
2
=
sin
6
π w
2
d
6
π w
2
d
2
=
sin
3
π w
d
3
π w
d
2
=
sin
3
π
(0
.
03 mm)
0
.
18 mm
3
π
(0
.
03 mm)
0
.
18 mm
2
=
0
.
405285
.
003
10.0 points
Using a conventional twoslit apparatus with
light of wavelength 593 nm, 29 bright fringes
per centimeter are observed on a screen 4 m
away.
What is the slit separation?
Correct answer: 6
.
8788 mm.
Explanation:
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Kapoor (mk9499) – homework38 – Turner – (60230)
2
Let :
N
= 29 cm
−
1
= 2900 m
−
1
,
λ
= 593 nm = 5
.
93
×
10
−
7
m
,
and
L
= 4 m
.
The distance on the screen to the
m
th
and
(
m
+ 1)
th
bright fringe is
Δ
y
= (
m
+ 1)
λ L
d

m
λ L
d
=
λ L
d
.
Then the number of fringes per unit length is
N
=
1
Δ
y
=
d
λ L
and the slit separation is
d
=
λ L N
= (5
.
93
×
10
−
7
m) (4 m) (2900 m
−
1
)
×
1000 mm
m
=
6
.
8788 mm
.
004
10.0 points
Two coherent microwave sources that produce
waves of wavelength 1 cm are in the
xy
plane,
one on the
y
axis at
y
= 9 cm and the other
at
x
= 3 cm,
y
= 8 cm.
If the sources are in phase, find the differ
ence in phase between the two waves from
these sources at the origin.
Correct answer: 164
.
159
◦
.
Explanation:
Let :
vectorx
1
= (
x
1
, y
1
) = (0
,
9 cm)
,
vectorx
2
= (
x
2
, y
2
) = (3 cm
,
8 cm)
,
and
λ
= 1 cm
.
The path difference between the distances
from these two sources to the origin is
Δ
r
=
r
1

r
2
=
radicalBig
x
2
1
+
y
2
1

radicalBig
x
2
2
+
y
2
2
= 9 cm

radicalBig
(3 cm)
2
+ (8 cm)
2
= 0
.
455996 cm
,
so the difference in phase is
δ
=
Δ
r
λ
360
◦
=
(0
.
455996 cm) (360
◦
)
1 cm
=
164
.
159
◦
.
005
10.0 points
Hint:
Use a small angle approximation;
e.g.
,
sin
θ
= tan
θ .
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 Turner
 Physics, Work, Diffraction, Light, Wavelength, Sin, Correct Answer, Kapoor

Click to edit the document details