Chapter 17 Reflection and Refraction

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Chapter 17Reflection and Refraction

• Law of Reflection
• The angle of incidence equals the angle of
  reflection

lti ltr
Regular Reflection
Diffuse Reflection
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Chapter 17Reflection and Refraction

• Refraction
• The bending of a wave as it enters a new medium

incident ray
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Chapter 17Reflection and Refraction

• Optical Density
• The property of the medium that determines the
  speed of light in a medium.
• Light bends toward the normal if the speed is
  reduced as it enters a new medium. (Less dense to
  more dense)
• Light bends away from the normal if the speed is
  increased as it enters a new medium. (More dense
  to less dense)

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Chapter 17Reflection and Refraction
Snells Law A ray of light bends in such a way
that the ratio of the the sine of the incident
ray to the sine of the refracted ray is a
constant.
i angle of incidence r angle of refraction n
optical density (pg 397) c speed of light in a
vacuum v speed of light in the medium
n1sin i n2 sin r n c/v
Illustration
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Chapter 17Reflection and Refraction
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Chapter 17Reflection and Refraction
A ray of light travels from the air into water at
an angle with the surface of 60º. Find the angle
of refraction. What is the speed of the wave in
the water?
n1sin i n2 sin r
lti 30º n1 1.00 n2 1.33
1 sin 30º 1.33 sin r
.5/1.33 sin r
n c/v
.376 sin r
v c/n 3 x 108 m/s / 1.33
r sin-1.376 22º
v 2.26 x 108 m/s
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Chapter 17Reflection and Refraction
A ray of light travels from the air into a solid
substance. The angle of incidence is 45º and the
angle of refraction is 17º.What is the medium
and what is the speed of the wave in the medium?
n1sin i n2 sin r
lti 45º ltr 17º n1 1.00 n2
1 sin 45º n2 sin 17º
.707 n2 .292
n c/v
n2 .707/.292
v c/n 3 x 108 m/s / 2.42
n2 2.42 diamond
v 1.24 x 108 m/s
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Chapter 17Reflection and Refraction
An insulated bulb is placed on the bottom of a
swimming pool 10 meters deep and 7 meters
from the wall. At what angle does the light leave
the pool?
i tan-1 7/10
n1sin i n2 sin r
i 35º
1.33 sin 35º 1.00 sin r
n1 1.33 n2 1.00
.763 sin r
r 50º
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Chapter 17Reflection and Refraction
Proof of Snells Law
Lines are drawn between the rays or direction of
travel of the light. The wavefronts or peaks of
the waves are perpendicular to the rays.
Remember that the electric and magnetic fields
(which are the wave material so to speak) are
perpendicular the direction of travel of the
waves.
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Chapter 17Reflection and Refraction
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Chapter 17Reflection and Refraction
The distances BC  and AD   correspond to equal
intervals of time T. The distances BC  and AD 
are different because light travels with
different speeds in the two optical substances or
media.                    
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Chapter 17Reflection and Refraction
AB is perpendicular to BC because AB is parallel
the wave front in the top medium.
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Chapter 17Reflection and Refraction
For triangle ABC
For triangle ACD
Divide left equation by Right equation
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Chapter 17Reflection and Refraction
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Chapter 17Reflection and Refraction
Critical Angle The incident angle that causes
the refracted ray to lie along the boundary of a
surface.
ic
total internal reflection
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Chapter 17Reflection and Refraction
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Chapter 17Reflection and Refraction

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Chapter 17Reflection and Refraction

• Find the critical angle between a water-diamond
• interface.

n1 2.42 n2 1.33
ic
n12.42
n1sin i n2 sin r
n2 1.33
2.42 sin ic 1.33 sin 90
sin ic 1.33/2.42
ic sin-1.550 33
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Chapter 17Reflection and Refraction

• Find the critical angle between a quartz-air
• interface.

n1 1.54 n2 1.00
ic
n11.54
n1sin i n2 sin r
n2 1.00
1.54 sin ic 1.00 sin 90
sin ic 1.00/1.54
ic sin-1.649 40
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Chapter 17Reflection and Refraction
Light travels from air into an optical fiber with
an index of refraction of 1.44.  (a)  In which
direction does the light bend?  (b)  If the
angle of incidence on the end of the fiber is
22o, what is the angle of refraction
inside the fiber?  (c)  Sketch the path of light
as it changes media
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Chapter 17Reflection and Refraction

• Since the light is traveling from a rarer region
• (lower n) to a denser region (higher n),
• it will bend toward the normal.
• (b) n1 sin i n2 sin r
• (1.00) sin 22o 1.44 sin r. sin r
  (1.00/1.44) sin 22o 0.260 r sin-1 (0.260)
  15o.
• (c)

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Chapter 17Reflection and Refraction
A ray of light in air is approaching the boundary
with a layer of crown glass at an angle of 42
degrees. Determine the angle of refraction of
the light ray upon entering the crown glass and
upon leaving the crown glass
Boundary 1 1.00 sin (42) 1.52 sin(r) 0.669
1.52 sin (r) 0.4402 sin (r) sin-1 (0.4402)
26.1 r 26.1
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Chapter 17Reflection and Refraction
A ray of light in air is approaching the boundary
with a layer of crown glass at an angle of 42
degrees. Determine the angle of refraction of
the light ray upon entering the crown glass and
upon leaving the crown glass
Boundary 2 1.52 sin (26.1) 1.00 sin(
r) 1.52 (0.4402) 1.00 sin ( r) 0.6691 sin
(r) sin-1 (0.6691) r r 42.0 degrees
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Chapter 17Reflection and Refraction
There is an important conceptual idea which is
found from an inspection of the above answer.
The ray of light approached the top surface of
the layer at 42 degrees and exited through the
bottom surface of the layer with the same angle
of 42 degrees. The light ray refracted one
direction upon entering and the other direction
upon exiting the two individual effects have
balanced each other and the ray is moving in the
same direction. The important concept is this
When light approaches a layer which has the
shape of a parallelogram that is bounded on both
sides by the same material, then the angle at
which the light enters the material is equal to
the angle at which light exits the layer.
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 The Secret of the Archer Fish In the quiet
waters of the Orient, there is an unusual fish
known as the Archer fish. The Archer fish is
unlike any other fish in that the Archer fish
finds its prey living outside the water. An
insect, butterfly, spider or similar creature is
the target of the Archer fish's powerful spray of
water. The Archer fish will search for prey that
is resting upon a branch or twig above the water.
The fish then positions itself underneath the
prey and with pinpoint accuracy knocks the prey
off the branch using a powerful jet of water. The
prey falls to the water, and the Archer fish
swims to the surface to retrieve its meal. The
feat of shooting a stream of water to knock the
prey off a branch is remarkable. The fact that
the Archer fish can do this time and again with
pinpoint accuracy is even more remarkable. But
most remarkable of all is that the Archer fish
can accomplish this trick despite the fact that
light from the target to its eye undergoes
refraction at the air-water boundary. Such
refraction would cause a visual distortion,
making the prey appear to be in a location where
it isn't. Yet the Archer fish is hardly ever
fooled. What is the secret of the Archer fish?

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There is only one condition in which light can
pass from one medium to another, change its
speed, and still not refract. If the light is
traveling in a direction which is perpendicular
to the boundary, no refraction occurs. As the
light wave crosses over the boundary, its speed
and wavelength still change. Yet, since the light
wave is approaching the boundary in a
perpendicular direction, each point on the
wavefront will reach the boundary at the same
time for this reason, there is no refraction of
the light. Such a ray of light is said to be
approaching the boundary while traveling along
the normal. (The normal is a line drawn
perpendicular to the surface.)


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Effects of Refraction
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