Version PREVIEW – HW 5 – Savrasov – (39824)
1
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printout
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have
11
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Bright Band Separation
001
(part 1 of 2) 10.0 points
Two slits are illuminated by a 533 nm light.
The angle between the zerothorder bright
band at the center of the screen and the
fourthorder bright band is 17
.
4
◦
.
If the screen is 167 cm from the doubleslit,
how far apart is this bright band from the
central peak?
Correct answer: 52
.
3346 cm.
Explanation:
Basic Concept:
y
=
L
tan
θ.
Solution:
y
=
L
tan
θ
= (167 cm) tan(17
.
4
◦
)
= 52
.
3346 cm
.
002
(part 2 of 2) 10.0 points
What is the distance between the two slits?
Correct answer: 0
.
00712946 mm.
Explanation:
Basic Concept:
d
sin
θ
=
mλ
Solution:
The fourthorder bright band oc
curs when
m
= 4; therefore,
d
=
m λ
sin
θ
=
4 (533 nm)
sin(17
.
4
◦
)
= 0
.
00712946 mm
.
Double Slit Wave Length 01
003
10.0 points
A screen is illuminated by monochromatic
light as shown in the figure below.
The distance from the slits to the screen is
5 m
.
1
.
9 cm
5 m
0
.
9 mm
S
1
S
2
θ
viewing
screen
What is the wave length if the distance
from the central bright region to the sixth
dark fringe is 1
.
9 cm
.
Correct answer: 621
.
814 nm.
Explanation:
Basic Concepts:
For bright fringes, we
have
d
sin
θ
=
m λ ,
and for dark fringes, we have
d
sin
θ
=
parenleftbigg
m
+
1
2
parenrightbigg
λ ,
where
m
= 0
,
±
1
,
±
2
,
±
3
,
· · ·
.
From geometry, we have
y
=
L
tan
θ .
Let
:
y
= 1
.
9 cm = 0
.
019 m
,
L
= 5 m
,
and
d
= 0
.
9 mm = 0
.
0009 m
.
r
2
r
1
y
L
d
S
1
S
2
θ
= tan
−
1
parenleftBig
y
L
parenrightBig
viewing
screen
δ
≈
d
sin
θ
≈
r
2

r
1
P
O
negationslash
S
2
Q S
1
≈
90
◦
Q
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Version PREVIEW – HW 5 – Savrasov – (39824)
2
r
2
r
1
d
S
1
S
2
θ
= tan
−
1
parenleftBig
y
L
parenrightBig
θ
δ
≈
d
sin
θ
≈
r
2

r
1
negationslash
S
2
Q S
1
≈
90
◦
Q
Solution:
The angle
θ
from the slits’ mid
point to the
y
position on the screen is
θ
= arctan
bracketleftBig
y
L
bracketrightBig
= arctan
bracketleftbigg
(0
.
019 m)
(5 m)
bracketrightbigg
= 0
.
00379998 rad
.
The wavelength of the light for the sixth dark
fringe,
m
= 5, is
λ
=
d
sin
θ
parenleftbigg
m
+
1
2
parenrightbigg
=
(0
.
0009 m) sin(0
.
00379998 rad)
(5
.
5)
= 6
.
21814
×
10
−
7
m
= 621
.
814 nm
.
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 Spring '09
 RichardScalettar
 Physics, Light, Wavelength, Doubleslit experiment, phase angle difference, δ θ, Version PREVIEW

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