1. 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

2. Chapter 17 Reflection and Refraction
The bending of a wave as it enters a new medium
incident ray r
refracted ray
reflected ray i

3. Chapter 17 Reflection 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)

4. 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.
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 r i
incident ray
refracted ray n1 n2
Illustration

5. Chapter 17 Reflection and Refraction

6. 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?
n1sin i = n2 sin r
<i = 30º
n1 =
n2 = 1.33
30º
incident ray
n1= 1.00
1 sin 30º = 1.33 sin r r
refracted ray
n2 = 1.33
.5/1.33 = sin r
n = c/v
.376 = sin r
v = c/n = 3 x 108 m/s / 1.33
r = sin = 22º
v = 2.26 x 108 m/s

7. 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?
n1sin i = n2 sin r
<i = 45º
<r = 17º
n1 =
n2 =
45º
incident ray
n1= 1.00
1 sin 45º = n2 sin 17º
17º
refracted ray
n2 =
.707 = n2 .292
n = c/v
n2 = .707/.292
v = c/n = 3 x 108 m/s / 2.42
n2 = diamond
v = 1.24 x 108 m/s

8. Chapter 17 Reflection and Refraction
10 m
7 m i r
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 = sin r
n1 = 1.33
n2 = 1.00
.763 = sin r
r = 50º

9. 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.

10. Chapter 17 Reflection and Refraction

11. 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.
                   

12. Chapter 17 Reflection and Refraction
AB is perpendicular to BC
because AB is parallel the
wave front in the top
medium.

13. Chapter 17 Reflection and Refraction
For triangle ABC
For triangle ACD
Divide left equation by
Right equation

14. Chapter 17 Reflection and Refraction

15. 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

16. Chapter 17 Reflection and Refraction

17. Chapter 17 Reflection and Refraction

18. Chapter 17 Reflection and Refraction
Find the critical angle between a water-diamond
interface. ic
n1 = 2.42
n2 = 1.33
n1=2.42
n1sin i = n2 sin r
n2= 1.33
2.42 sin ic = 1.33 sin 90°
sin ic = 1.33/2.42
ic = sin = 33°

19. Chapter 17 Reflection and Refraction
Find the critical angle between a quartz-air
interface. ic
n1 = 1.54
n2 = 1.00
n1=1.54
n1sin i = n2 sin r
n2= 1.00
1.54 sin ic = 1.00 sin 90°
sin ic = 1.00/1.54
ic = sin = 40°

20. 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

21. Chapter 17 Reflection 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 = r = sin-1 (0.260) = 15o.
(c)

22. 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) = sin(r)
0.669 = sin (r)
= sin (r)
sin-1 (0.4402) = 26.1
r =26.1

23. 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)
= sin (r)
sin-1 (0.6691) = r
r = 42.0 degrees

24. 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.

25. 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?

26. 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.)

27. 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.
                                                             

29. Effects of Refraction