01.
02.
03.
Physics 2220 – Module 12 Homework
The mirror in the figure deflects a horizontal laser beam by 60 o ϕ
This is an exercise in geometry. Notice:
θ r
θ i
+ ϕ =
+ ( 60 o
90 o
− ϕ) = 90 o
Since the incident and reflected are the same, set the equations equal.
90 o
− ϕ = 90 o
60 o
= 2
− ϕ
60 o ϕ = 30 o
+ ϕ
An underwater diver sees the sun 50 o above the horizontal. How high is the Sun above the horizontal to a fisherman in the boat above the diver?
Use Snell's Law to determine the incident angle of the
Sun's rays.
n air sin (θ air
) = n water
θ air
= sin − 1
( n water n air sin (θ water
) sin (θ water
)
)
θ air
= sin − 1
( 1.33
1.00
sin ( 40 o
)
)
= 58.7
o
So the angle someone above the water would see the
Sun above the horizon:
θ = 90 o
− 58.7
o
= 31.3
o
It's nighttime, and you've dropped your goggles into a 3.0-m-deep swimming pool. If you hold a laser pointer 1.0 m above the edge of the pool, you can illuminated the goggles if the laser beam enters the water 2.0 m from the edge. How far are the goggles from the edge of the pool?
Use some geometry to find the incident angle:
θ air
= tan − 1
( 2.0 m
1.0
m
)
= 63.43
o
Now get the refracted angle using Snell's Law: n air sin (θ air
) = n water sin (θ water
)
θ water
= sin − 1
( n air n water sin (θ air
)
)
θ airwater
= sin − 1
( 1.00
1.33
sin ( 63.43
o
)
)
= 42.26
o
Use geometry to find x.
tan (θ water
) = x
3.0 m
→ x = ( 3.0 m ) ( tan ( 42.26
o
)) = 2.73 m
Total Distance is therefore 4.73 m .

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05.
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The glass core of an optical fiber has an index of refraction of 1.60. The index of refraction of the cladding is 1.48. What is the maximum angle a light ray can make with the wall of the core if it is to remain inside the fiber?
Simply determine the critical angle.
θ c
= sin − 1
( n cladding n core
)
= sin − 1
( 1.48
1.60
)
= 67.7
o
A beam of light strikes a sheet of glass at an angle of 57.0º with the normal in air. The red light makes an angle of 38.1º with the normal in the glass, while the violet light makes an angle of 36.7º.
(a)
(b)
What are the indexes of refraction of this glass for these colors of light?
Use Snell's law to solve for the indexes of refraction: n air sin (θ air
) = n glass sin (θ glass
)
Red: n glass
= n air sin (θ air sin (θ glass
)
)
=
( 1 ) sin ( 57.0
ο
) sin ( 38.1
ο
)
= 1.36
Violet: n glass
= n air sin (θ air sin (θ glass
)
)
=
( 1 ) sin ( 57.0
ο
) sin ( 36.7
ο
)
= 1.40
What are the speeds of red and violet light in the glass?
Now that we have the indexes of refraction, speeds can be found.
Red: v = c n
=
3.00
× 10 8
1.36
m/s
= 2.21
× 10 8 m/s
Violet: v = c n
=
3.00
× 10
8
1.40
m/s
= 2.14
× 10 8 m/s
A sheet of glass has n
30 o red
= 1.52 and n violet
= 1.55. A narrow beam of white light is incident on the glass at
. What is the angular spread of the light inside the glass?
Use Snell's Law to determine the refracted angle for each of the wavelengths.
θ red
= sin − 1
( n
1 n red n
1 sin (θ
1
) = n red sin (θ red
) sin (θ
1
)
)
= sin − 1
( 1.00
1.52
sin ( 30 o
)
)
= 19.204897
o
θ blue
= sin − 1
( n
1 n blue n
1 sin (θ
1
) = n blue sin (θ blue
) sin (θ
1
)
)
= sin − 1
( 1.00
1.55
sin ( 30 o
)
)
= 18.8190633
o
The difference between the refracted angles is the angular spread.
δ = θ red
− θ violet
= 0.4
o

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08.
Only 25% of the intensity of a polarized light wave passes through a polarizing filter. What is the angle between the electric field and the axis of the filter?
Use the relationship between the initial and final intensity of light:
I = I o cos 2
(θ)
I
I o
= 0.25
= cos 2
(θ) cos (θ) =
θ = cos − 1
(
0.25
0.25
) = 60 o
A 200 mW vertically polarized laser beam passes through a polarizing filter whose axis is 35 horizontal. What is the power of the laser beam as it emerges from the filter?
o from
Start with the relationship between the initial and final intensity of light.
I = I o cos 2
(θ)
The equation for wave intensity can apply to both the initial and transmitted intensity:
I =
P
A
I o
=
P
A o
Assuming the area of the beam does not change, the first equation becomes:
P = P o cos 2
(θ)
Solve for the power of the beam as it emerges from the filter. Since the filter's axis is 35 degrees from the horizontal, the angle between the electric field and the filter axis = 90 – 35 = 55 degrees
P = ( 200 × 10 − 3 W ) cos 2
( 55 o
) = 66 mW