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Chapter 35 Worksheet: Interference 1
Wavelength & Index of Refraction:
- The wave speed for 590 nm light is 2.25x
8
m/s in water.
a) What is the index of refraction for water at this wavelength?
Ans.
8 m
s
8 m
s
3.0x c
n = = = 1.
v 2.25x
b) If 590 nm light has an incident angle of 30
o
onto a settled surface of water, what is the
angle of the refracted light?
Ans. Re-arranging Snell’s Law:
o
-1 o
1.0 sin
θ = sin = 22.
c) If 590 nm light were shined up through water toward the surface, at what angle would all
of the light be reflected and not transmitted into the air? This is referred to as the critical
angle for total internal reflection.
Ans. The condition for total internal reflection is
o
-1 o
C
1.0 sin
θ = sin = 48.
- A monochromatic light source (λ=450 nm) is shined onto a transparent material (n = 1.5).
a) How fast does the light travel through this material?
Ans.
8 m
8 s m
s
3.0x c
v = = = 2.0x
n 1.
b) What is the frequency of the light in this material?
Ans.
8 m
14 s
air
c 3.0x
f = = = 6.7x10 Hz
λ 4.50x10 m
c) What is the wavelength of light in this material?
Ans.
air
material
4.50x10 m
= = = 3.0x10 m
n 1.
Young’s Experiment (Two Slit Interference):
Two parallel slits (width = 1.0x
m) are side-by-side and separated by a distance of 1.5 x
10
m. A monochromatic light source with a wavelength of 600 nm is transmitted through
the slits.
a) Determine the angular location of the 1
st
interference maxima from the center.
Ans.
-7 m
-1 o s
6.0x
θ = sin = 0.
1.6x10 m
Note: the center to center distance is 1.6x
m.
b) A screen is placed 2.0 m directly in front of the slits. What is the distance from the center
of the central maxima to the 1
st
interference maxima?
Ans.
o
y = (2m) tan(0.21 ) = 0.0073m⋅
c) What is the distance between the 1
st
and 2
nd
maxima?
Ans.
o o
y = (2m) tan(0.43 ) - tan(0.21 ) = 0.0077m
Intensity for Two Slit Interference:
Chapter 35 Worksheet: Interference 2
- In a double slit experiment, a viewing screen is located at L=4.00 m from the slits. The
wavelength for the light source is 500 nm and the separation distance between the slits is
1.5x
m.
a) How far away (in m) from the center of the central intensity maxima is the m=2 maxima?
Ans.
2(5.00x10 m)
y = (4.00m)tan sin = 0.027m
1.5x10 m
b) Calculate the ratio of the intensities, I m=
to I
m=
.
Ans.
2 2
o m=
m=
2 2 o
m=
o m=
2
m=
2
m=
d d 2λ
4I cos sin θ cos
I
λ λ d
d d I
4I cos sin θ cos sin 0
λ λ
cos 2 I
I cos 0
Diffraction:
A diffraction grating has a spacing of 1.5 x 10
m. A monochromatic light source with a
wavelength of 500 nm is transmitted through the grating.
a) Determine the angular location of the 1
st
interference maxima.
Ans.
-7 m
-1 o s
5.0x
θ = sin = 19.
1.5x10 m
b) If the grating is located 2.5 m from the screen, what is the distance from the center of the
central maximum to the 1
st
interference maximum?
Ans.
o
y = (2.5m)tan 19.5 = 0.89m
Consider 2 diffraction gratings (spacing = 1.0 x 10
m and 2.0 x 10
m).
a) Which grating will produce the greatest angular separation between its maxima?
Ans. The smaller grating spacing will produce the larger the angular separation.
b) What is the angular separation between the 1
st
maxima produced by the two gratings?
The light source has a wavelength of 500 nm.
Ans.
-7 -
-1 -1 -1 -1 o
-6 -
1 2
5.00x10 m 5.00x10 m
θ = sin - sin = sin - sin = 15.
d d 1.0x10 m 2.0x10 m
c) If both gratings are located 2.0 m from the screen, what is the distance between the 2
maxima (on the same side of the central maxima?
Ans.
o
∆ d = L tan( ⋅ ∆θ) = 2m ⋅tan 15.5 = 0.55 m
Chapter 35 Worksheet: Interference 4
- Consider a light wave traveling through water (n water
= 1.33 and λ
air
= 600 nm) incident on a
thin layer of glass (n glass
= 1.5). The glass layer is completely submerged in water.
a. What is the minimum thickness of the glass that will result in destructive interference?
Ans. This problem is treated the same as the above problem, except that the incident wave
travels in water not air. The wavelength of the wave in water is: λ water
= λ air
/n water
water air
min
glass water glass
λ λ 6.00x10 m
d = = = = 1.50x10 m
2n 2n n 2 (1.33)(1.5)⋅
b. What is the minimum thickness of the glass that will result in constructive interference?
Ans.
water air -
min
glass water glass
λ λ 6.00x10 m
d = = = = 7.50x10 m
4n 4n n 4 (1.33)(1.5)⋅
c. If the glass is 300 nm in thickness, which wavelength(s) of visible light, and corresponding
m value(s), will result in destructive interference?
Ans.
min water glass
air
2(3.00x10 m)(1.33)(1.5) 2d n n
1.20x10 m
λ = = =
m m m
The wavelength range for visible light is roughly 400 nm < λ
air
< 700 nm,
m = 2: λ
air
= 599 nm
m = 3: λ air
= 400 nm
d. If the glass is 300 nm in thickness, which wavelength(s) of visible light, and corresponding
m value(s), will result in constructive interference?
Ans.
min water glass
air
1 1
2 2
2(3.00x10 m)(1.33)(1.5) 2d n n
1.20x10 m
λ = = =
m m + m +
The wavelength range for visible light is roughly 400 nm < λ
air
< 700 nm,
m = 2: λ air
= 480 nm
