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optics
Ch 7 - Lecture 2
Dr. Haya Alhummiany
Phys 201
Refraction
Refraction
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Light rays may bend as they cross a boundary from one material to another, like
from air to water.
This bending of light rays is known as refraction.
When a ray of light crosses from one material to another, the amount it bends
depends on the difference in index of refraction between the two materials.
Index of Refraction, n
n = Index of Refraction
c = Speed of light in vacuum
v = Speed of light in medium
Note that a large index of refraction corresponds to a relatively
slow light speed in that medium.
Index of refraction
The ability of a material to bend rays of light is described by the
index of refraction (n).
Speed of Light & Refraction
As you have already learned, light is extremely fast, about
c = 3  108 m/s in a vacuum. Light, however, is slowed down by
the presence of matter. The extent to which this occurs depends
on what the light is traveling through. Light travels at about 3/4
of its vacuum speed (0.75 c ) in water and about 2/3 its vacuum
speed (0.67 c ) in glass.
The reason for this slowing is because when light strikes an
atom it must interact with its electron cloud. If light travels from
one medium to another, and if the speeds in these media differ,
then light is subject to refraction (a changing of direction at the
interface).
Speed of Light & Refraction
The
speed
at
which
light
propagates
through transparent materials, such as glass or air, is
less than c. The ratio between c and the speed v at
which light travels in a material is called the refractive
index n of the material (n = c / v). For example,
for visible light the refractive index of glass is typically
around 1.5, meaning that light in glass travels at c / 1.5
≈ 200000 km/s; the refractive index of air for visible light
is 1.000293, so the speed of light in air is 299705 km/s
or about 88 km/s slower than c.
Reflection & Refraction
At an interface between two media, both reflection and refraction can occur. The
angles of incidence, reflection, and refraction are all measured with respect to the
normal. The angles of incidence and reflection are always the same.
If light speeds up upon entering a new medium, the angle of refraction, r , will be
greater than the angle of incidence, as depicted on the left. If the light slows down
in the new medium, r will be less than the angle of incidence, as shown on the
right.
normal
normal
r
r
Snell's law of refraction
• Snell’s law is the relationship between the angles of
incidence and refraction and the index of refraction of both
materials.
Angle of incidence
(degrees)
Angle of refraction
(degrees)
ni sin Qi = nr sin Qr
Index of
refraction of
incident
material
Index of
refraction of
refractive
material
Calculate the angle of refraction
• A ray of light traveling
through air is incident on a
smooth surface of water at
an angle of 30° to the
normal.
• Calculate the angle of
refraction for the ray as it
enters the water.
Calculate the angle of refraction
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1) You are asked for the angle of refraction.
2) You are told the ray goes from air into water at 30 degrees.
3) Snell’s law:
ni sin(θi) = nr sin(θr)
ni = 1.00 (air), nr = 1.33 (water)
4) Apply Snell’s law to find θr.
1.00sin(30°) = 1.33 sin(θr)
sin(θr) = 0.5 ÷ 1.33 = 0.376
Use the inverse sine function to find the angle that has a
sine of 0.376.
θr = sin-1(0.376) = 22°
Critical Angle
nr
The incident angle that causes the
refracted ray to skim right along the
boundary of a substance is known
as the critical angle, c. The critical
angle is the angle of incidence that
produces an angle of refraction of
90º. If the angle of incidence
exceeds the critical angle, the ray is
completely reflected and does not
enter the new medium. A critical
angle only exists when light is
attempting to penetrate a medium
of higher optical density than it is
currently traveling in.
ni
c
From Snell,
n1 sinc = n2 sin 90
Since sin 90 = 1, we
have n1 sinc = n2 and
the critical angle is
c = sin-1
nr
ni
Critical Angle Sample Problem
Calculate the critical angle for the diamond-air boundary.
Refer to the Index of Refraction chart for the information.
c = sin-1 (nr / ni)
air
diamond
c
= sin-1 (1 / 2.42)
= 24.4
Any light shone on this
boundary beyond this angle
will be reflected back into the
diamond.