Download Chapter 23: Reflection and Refraction of Light

Chapter 23: Reflection and
Refraction of Light
•Huygens’s Principle
•Reflection
•Refraction
•Total Internal Reflection
•Polarization by Reflection
•Formation of Images
•Plane Mirrors
•Spherical Mirrors
•Thin Lenses
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§23.1 Huygens’s Principle
A set of points with equal phase is called a wavefront.
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A ray points in the direction of wave propagation and is
perpendicular to the wavefronts. Or a ray is a line in the
direction along which light energy is flowing.
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Huygens’s principle: At some time t, consider every point
on a wavefront as a source of a new spherical wave. These
wavelets move outward at the same speed as the original
wave. At a later time t+t, each wavelet has a radius vt,
where v is the speed of propagation of the wave. The
wavefront at t+t is a surface tangent to the wavelets.
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Geometric optics is an approximation to the behavior of
light that applies when interference and diffraction are
negligible. In order for diffraction to be negligible, the sizes
of objects must be large compared to the wavelength of
light.
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§23.2 Reflection of Light
When light is reflected from a smooth surface the rays
incident at a given angle are reflected at the same angle.
This is specular reflection.
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Reflection from a rough surface is called diffuse reflection.
“Smooth” and “rough” are determined based on the
wavelength of the incident rays.
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The angle of incidence equals the angle of reflection. The
incident ray, reflected ray, and normal all lie in the same
plane. The incident ray and reflected ray are on opposite
sides of the normal.
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§23.3 Refraction of Light
When light rays pass from one medium to another they
change direction. This is called refraction.
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Snell’s Law
n1 sin 1  n2 sin 2
where the subscripts refer
to the two different media.
The angles are measured
from the normal.
When going from high n to low n, the ray will bend away
from the normal.
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The incident ray, transmitted ray, and normal all lie in the
same plane.
The incident and transmitted rays are on opposite sides of
the normal.
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Example (text problem 23.11): Sunlight strikes the surface of
a lake. A diver sees the Sun at an angle of 42.0° with
respect to the vertical. What angle do the Sun’s rays in air
make with the vertical?
incident wave
1
n1 = 1.00; air
surface
n2 = 1.33; water
42°
Transmitted
wave
n1 sin 1  n2 sin  2
1.00sin 1  1.333sin 42
Normal
sin 1  0.8920
1  63.1
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§23.4 Total Internal Reflection
The angle of incidence for when the angle of refraction is
90° is called the critical angle.
n1 sin 1  n2 sin  2
n1 sin  c  n2 sin 90  n2
n2
sin  c 
n1
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If the angle of incidence is greater than or equal to the
critical angle, then no wave is transmitted into the other
medium. The wave is completely reflected from the
boundary.
Total internal reflection can only occur when the incident
medium has a larger index of refraction than the second
medium.
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Example (text problem 23.22): Calculate the critical angle
for sapphire surrounded by air.
2=90
n2 = 1.0; air
n1 = 1.77; sapphire
surface
1
incident wave
Transmitted wave
Normal
n1 sin 1  n2 sin  2
1.77 sin  c  1.00sin 90
sin  c  0.565
1  34.4
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§23.5 Polarization by Reflection
Brewster’s angle is the angle of incidence for which the
reflected light is completely polarized.
Light is totally polarized when the reflected ray and the
transmitted ray are perpendicular.
ni sin  i  nt sin  t
ni sin  B  nt sin 90   B   nt cos  B
nt
tan  B 
ni
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Example (text problem 23.32): (a) Sunlight reflected from
the still surface of a lake is totally polarized when the
incident light is at what angle with respect to the
horizontal?
nwater 1.33
tan  B 

 1.33
nair
1.00
 B  53.1
The angle is measured from the normal, so 90 - 53.1
= 36.9 is the angle from the horizontal.
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Example continued:
(b) In what direction is the reflected light polarized?
It is polarized
perpendicular
to the plane of
incidence.
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Example continued:
(c) Is any light incident at this angle transmitted into the
water? If so, at what angle below the horizontal does the
transmitted light travel?
From Snell’s Law:
n1 sin 1  n2 sin  2
1.00sin 53.1  1.333sin  2
sin  2  0.6000
 2  36.9
The angle is measured from the normal, so 90 - 36.9
= 53.1 is the angle from the horizontal.
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§23.6 Formation of Images
An image is real if light rays from a point on the object
converge to a corresponding point on the image.
A camera lens
forms a real image.
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The light rays
appear to come
from behind the
mirror.
Your eye focuses
the diverging rays
reflected by the
mirror.
An image is virtual if the light
rays from a point on the object
are directed as if they diverged
from a point on the image, even
though the rays do not actually
pass through the image point.
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Example (text problem 23.35): A defect in a diamond
appears to be 2.00 mm below the surface when viewed from
directly above that surface. How far beneath the surface is
the defect.
Air
n2 =1.00
2
2
Surface
Diamond
n1 = 2.419
1
1
y’
y
Actual location
of defect
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Example continued:
The angles 1 and 2 are related by Snell’s Law:
n1 sin 1  n2 sin 2
The actual depth of the defect is y and it appears to be at a
depth of y’. These quantities are related by:
y tan 2  y tan 1
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Example continued:
Dividing the previous two expressions gives:
n1 y cos1  n2 y cos2
As long as you are directly above the defect and its image,
the angles 1 and 2 are nearly 0°. Rays from only a narrow
range of angles will enter your eye. The above expression
simplifies to:
n1 y  n2 y
y  n2

y n1
(general result)
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Example continued:
The actual depth of the defect in the diamond is then
n1
 2.419 
y  y  
2.00 mm   4.84 mm.
n2
 1.00 
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§23.7 Plane Mirrors
A point source and its image are at the same distance from
the mirror, but on opposite sides of the mirror.
Treat an extended
object as a set of
point sources.
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Example (text problem 23.41): Entering a darkened room,
Gustav strikes a match in an attempt to see his
surroundings. At once he sees what looks like another
match about 4 m away from him. As it turns out, a mirror
hangs on one of the walls. How far is Gustav from the wall
with the mirror?
The image seems 4 m away, but the mirror is only 2 m
away since the rays will appear to come from a point 2 m
behind the mirror.
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§23.8 Spherical Mirrors
vertex
Center of
curvature
Principal
axis
The focal
point
A convex (or diverging) mirror curves
away from the observer.
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A ray parallel to
the principle axis
is reflected, and it
appears to have
come from point
F, the focal point
of the mirror.
For a convex mirror, the focal point is on the axis and is
located a distance 0.5R behind the mirror, where R is the
radius of curvature.
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Drawn in green, red, and blue are the principal rays.
1. A ray parallel to the principal axis is reflected as if it came
from the focal point. (green)
2. A ray along a radius is reflected back upon itself. (red)
3. A ray directed toward the focal point is reflected parallel to
the principal axis. (blue)
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For the pencil in the previous figure, the image is upright,
virtual, smaller than the object, and closer to the mirror
than the object.
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A concave (or converging) mirror
curves toward the observer.
Center of
curvature
vertex
Principal
axis
The focal
point
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Drawn in green, red, and blue are the principal rays.
1. A ray parallel to the principal axis is reflected through the
focal point. (green)
2. A ray along a radius is reflected back upon itself. (red)
3. A ray along the direction from the focal point to the mirror is
reflected parallel to the principal axis. (blue)
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The magnification is defined as
image size h
m
 .
object size h
An inverted image has m<0 and an upright image has
m>0.
The expression for magnification can also be written as
m
q
p
where p is the object distance and
q is the image distance.
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The mirror equation:
1 1 1
 
p q f
where f is the focal length of the mirror.
f<0 when the focal point is behind the
mirror.
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Example (text problem 23.46): An object 2.00 cm high is
placed 12.0 cm in front of a convex mirror with a radius of
curvature of 8.00 cm. Where is the image formed?
1 1 1
 
p q f
where p = 12.0 cm, f = -0.5R = -4.00 cm, and q is the
unknown image distance. Solving gives q = -3.00 cm. The
image is behind the mirror.
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§23.9 Thin Lenses
A diverging lens will bend light away from the principle axis.
A converging lens will bend light toward the principal axis.
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Magnification:
h
q
m 
h
p
The thin lens equation:
1 1 1
 
p q f
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Example (text problem 23.64): A diverging lens has a
focal length -8.00 cm.
(a) What are the image distances for objects placed at
various distances from the lens? Is the image real or
virtual? Upright or inverted? Enlarged or diminished?
Object
distance
Image
distance
Real /
virtual?
Upright /
inverted?
Enlarged/
diminished
5 cm
-3.08 cm
Virtual
upright
Diminished
8 cm
-4.00 cm
Virtual
upright
Diminished
14 cm
-5.09 cm
Virtual
upright
Diminished
16 cm
-5.33 cm
Virtual
upright
Diminished
20 cm
-5.71 cm
Virtual
upright
Diminished
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Example continued:
(b) If the object is 4.00 cm high, what is the height of the
image?
Object
distance
Image
distance
Magnification
Image height
5 cm
-3.08 cm
0.616
2.46 cm
8 cm
-4.00 cm
0.500
2.00 cm
14 cm
-5.09 cm
0.364
1.45 cm
16 cm
-5.33 cm
0.333
1.33 cm
20 cm
-5.71 cm
0.285
1.14 cm
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Summary
•The Laws of Reflection
•The Laws of Refraction
•Condition for Total Internal Reflection
•Condition for Total Polarization of Reflected Light
•Real/virtual Images
•Mirrors (plane & spherical)
•Thin Lenses
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