Download Chapter 17 Reflection and Refraction

Chapter 17
Reflection and Refraction

Law of Reflection
The angle of incidence equals the angle of
reflection
i r
<i = <r
Regular Reflection
Diffuse Reflection
Chapter 17
Reflection and Refraction

Refraction
The bending of a wave as it enters a
new medium
incident ray
reflected ray
i r
r
refracted ray
Chapter 17
Reflection and Refraction

Optical Density
The property of the medium that determines the
speed of light in a medium.
1.
Light bends toward the normal if the speed is
reduced as it enters a new medium. (Less dense
to more dense)
2.
Light bends away from the normal if the speed
is increased as it enters a new medium. (More
dense to less dense)
Chapter 17
Reflection and Refraction
Snell’s 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.
n1sin i = n2 sin r
n = c/v
incident ray
i
r
n1
n2
refracted ray
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
Illustration
Chapter 17
Reflection and Refraction
Chapter 17
Reflection 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?
incident ray
n1sin i = n2 sin r
<i = 30º
30º
n1 = 1.00
n1= 1.00
1 sin 30º = 1.33 sin r n2 = 1.33
.5/1.33 = sin r
.376 = sin r
r = sin-1.376 = 22º
n = c/v
r
v = c/n = 3 x 108 m/s / 1.33
v = 2.26 x 108 m/s
n2 = 1.33
refracted ray
Chapter 17
Reflection 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?
incident ray
n1sin i = n2 sin r
<i = 45º
45º
<r = 17º
n1= 1.00
1 sin 45º = n2 sin 17º n1 = 1.00
n2 =
n2 =
.707 = n2 .292
17º
n = c/v
refracted ray
n2 = .707/.292
v = c/n = 3 x 108 m/s / 2.42
n2 = 2.42 diamond
v = 1.24 x 108 m/s
Chapter 17
Reflection 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?
n1sin i = n2 sin r
1.33 sin 35º = 1.00 sin r
.763 = sin r
r = 50º
r
10 m
i = tan-1 7/10
i = 35º
n1 = 1.33
n2 = 1.00
i
7m
Chapter 17
Reflection and Refraction
Proof of Snell’s 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.
Chapter 17
Reflection and Refraction
Chapter 17
Reflection 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.
Chapter 17
Reflection and Refraction
AB is perpendicular to BC
because AB is parallel the
wave front in the top
medium.
Chapter 17
Reflection and Refraction
For triangle ABC
Divide left equation by
Right equation
For triangle ACD
Chapter 17
Reflection and Refraction
Chapter 17
Reflection and Refraction
Critical Angle The incident angle that causes the refracted ray
to lie along the boundary of a surface.
air
water
ic
total internal reflection
Chapter 17
Reflection and Refraction
Chapter 17
Reflection and Refraction
Chapter 17
Reflection and Refraction
Find the critical angle between a water-diamond
interface.
ic
n1=2.42
n2= 1.33
n1 = 2.42
n2 = 1.33
n1sin i = n2 sin r
2.42 sin ic = 1.33 sin
90°
sin ic = 1.33/2.42
ic = sin-1.550 = 33°
Chapter 17
Reflection and Refraction
Find the critical angle between a quartz-air
interface.
n1 = 1.54
ic
n2 = 1.00
n =1.54
1
n2= 1.00
n1sin i = n2 sin r
1.54 sin ic = 1.00 sin
90°
sin ic = 1.00/1.54
ic = sin-1.649 = 40°
Chapter 17
Reflection 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
Chapter 17
Reflection and Refraction
(a) 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)
Chapter 17
Reflection 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
Chapter 17
Reflection 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
Chapter 17
Reflection 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.
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?
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.)
The secret to the Archer fish's success is that it
lines up its sight with the prey from a position
directly underneath the prey. From this vantage
point, light from the prey travels directly to the
fish's eye without undergoing a change in
direction. Since the light is traveling along the
normal to the surface, it does not refract; the light
passes straight through the water to the fish's
eyes. Normally, when light from an object
changes medium on the way to the eye, there is a
visual distortion of the image. But as the Archer
fish sights along the normal, there is no refraction
and no no visual distortion of the image. From
this ideal position, the Archer fish is able to hit its
prey time after time. The secret of the Archer fish
is to use its understanding of the physics of
refraction of light. The Archer fish knows that
refraction is less when sighting along the normal.
Now that's physics for better living. Like all fish,
the Archer fish has spent its life living in schools;
and there's no better place than a school to learn
about the physics of refraction.
Effects of Refraction