This document discusses various theories and models of light, including evidence that light behaves as both a wave and particle. It explains Young's double-slit experiment, which showed light can interfere like a wave. However, some experiments like the photoelectric effect are better explained by thinking of light as particles called photons. The document also discusses Snell's law of refraction, how light refracts when passing from one medium to another, total internal reflection, and applications of fiber optics.

1. Physics III
Part one
Optics
DR. EHAB HEGAZY

2. Is light wave or particle?
 Evidence for light as a wave model
 Young experiments show that light can interfere, if two
sources emitting waves it produce a pattern of
alternating bright and dark bands on a screen.
 Draw back here is no medium to support light wave
2

3. Particle theory (Newton's theory)
 Light is composed of corpuscles (particles of matter) which were emitted in
all directions from a source
 Newton's theory could be used to predict the reflection of light, but could only
explain refraction
3

4. Light is stream of particles
 Some experiments can not be explained by wave model
and can be explained by particle model done by Newton.
Such as photoelectric effect.
4

5. Quantization Model
 Einstein proposed that energy of light wave is present in particles
called photon
 According to Einstein theory:
E = hf
E energy of photon F is frequency electromagnetic wave.
h is plank constant .
5

6. Features of a wave
 Light can be propagated in vacuum.
 Light is transverse wave.
 Light propagates as simple harmonic motion (S.H..M)
6

7. Longitudinal Wave
wave particles vibrate back
and forth along the path that
the wave travels.
Compressional Wave

8.  Compressions
The close together part of the wave
 Rarefactions
The spread-out parts of a wave

9. Transverse waves
wave particles vibrate
in an up-and-down motion.

10. Transverse waves
 Crests
Highest part of a wave
 Troughs
The low points of the wave

11. Index of Refraction, n
The index of refraction of a substance is the ratio of the speed in light in a vacuum to
the speed of light in that substance:
n = Index of Refraction
c = Speed of light in vacuum
v = Speed of light in medium
n = c /v
Note that a large index of refraction
corresponds to a relatively slow light speed
in that medium.
Medium
Vacuum
Air (STP)
Water (20º C)
Ethanol
Glass
Diamond
n
1
1.00029
1.33
1.36
~1.5
2.42

12. 12

13. Snell’s Law
Snell’s law states that a ray of light bends in such a way
that the ratio of the sine of the angle of incidence to the
sine of the angle of refraction is constant.
Mathematically,
ni sin i = nr sinr
Here ni is the index of refraction in the original medium
and nr is the index in the medium the light enters. i and
r are the angles of incidence and refraction, respectively.
i
r
ni
nr
Snell

14. Snell’s Law Derivation Two parallel rays are shown. Points A
and B are directly opposite one
another. The top pair is at one point in
time, and the bottom pair after time t.
The dashed lines connecting the pairs
are perpendicular to the rays. In time
t, point A travels a distance x, while
point B travels a distance y.
sin1 = x / d, so x = d sin1
sin2 = y / d, so y = d sin2
Speed of A: v1 = x / t
Speed of B: v2 = y / t
Continued…
•
••
•
A
A B
B
1
2
x
y
d
n1
n2

15. Snell’s Law Derivation
(cont.)
v1 /c sin1 1/n1 sin1 n2
v2 /c sin2 1/n2 sin2 n1
=  = =
 n1 sin1 = n2 sin2
v1 x/ t x sin1
=
v2 y/ t y sin2
= = So,
•
••
•
A
A B
B
1
2
x
y
d
n1
n2

17. Refraction Problem #1
1. Find the first angle of refraction using
Snell’s law.
2. Find angle ø. (Hint: Use Geometry
skills.)
3. Find the second angle of incidence.
4. Find the second angle of refraction, ,
using Snell’s Law
19.4712º

Glass, n2 = 1.5
Air, n1 = 1
30°
ø
79.4712º
10.5288º
Horiz. ray, parallel to
base
15.9º
Goal: Find the angular displacement of the ray after having passed through the
prism. Hints:

18. Refraction Problem #2
120º

d
glass
H20
H20
10m
20º 20º 0.504 m 5.2 ·10-8 s 26.4º
n1 = 1.3
n2 = 1.5
Goal: Find the distance the light ray displaced due to the thick window and how
much time it spends in the glass. Some hints are given.
1. Find 1 (just for fun).
2. To show incoming & outgoing
rays are parallel, find .
3. Find d.
4. Find the time the light spends in
the glass.
Extra practice: Find  if bottom medium is
replaced with air.

19. Refraction Problem #3
 = ?
36°
Goal: Find the exit angle relative to the horizontal.
19.8°
glass
air
The triangle is isosceles.
Incident ray is horizontal, parallel to
the base.
 =

20. 20

21.  A light ray of wavelength 589 nm traveling through air is incident on a
smooth, flat slab of crown glass at an angle of 30.0° to the normal, as
sketched in Figure Find the angle of refraction.
21

22. 22

23.  A light beam passes from medium 1 to medium 2, with the
latter medium being a thick slab of material whose index of
refraction is n2 Show that the emerging beam
is parallel to the incident beam.
23

24. Total internal reflection

25. Optical fibers

26. The operation of optical fibers
 Optical fibers are narrow tubes of glass fibers with a plastic coating that carry
light from one end to the other.
 The light bounces off the walls of the fiber and can even bounce around
corners. The properties of optical fibers make them useful for a wide range of
applications including:
 Medical - to transmit pictures of organs and arteries
 Industrial - to transmit pictures of the inside of complex machinery
 Communications - to transmit data over long distances without transmission
loss

27.  Light rays use total internal reflection to travel along the
fibers. In order for this to be achieved, the light ray must
hit the walls of the fiber at a minimum angle of 82°,
which is the critical angle for light travelling from glass
to plastic. Since the fibers are very narrow, this is
usually not a problem

28. Fiber optic & copper cable
Fiber optic cable has many advantages over copper cable.
 Fiber transmits data much faster over longer distances than copper.
 Fiber cable is also smaller diameter and weighs less than its copper counterpart, making it
ideal for a variety of cabling solutions.
 Fiber optics are immune to RFI (radio frequency interference) and EMI (Electromagnetic
Interference) making them ideal for applications where close proximity to electronic devices
can cause RFI and EMI disruption.
 Fiber optic cabling uses less power
 provides less signal degradation than copper cables.
 They are generally non-flammable, virtually unable to be tapped, and are better suited for
data and illumination transmission.

29. Numerical aperture

30. 30

32. LIGHT
Total Internal Reflection in Prisms

33. LIGHT
Total Internal Reflection in PrismsThank you