Reflection from transparent materials (Chapt. 33 last part)
When unpolarized light strikes a transparent surface like glass there is
both transmission and reflection, obeying Snell’s law and the law of
reflection, respectively,
θi
θr
θt
For glass (n ~ 1.5) at normal incidence (θi = 0) the reflected ray is ~ 4% of
the incident intensity. As the angle of incidence increases more light is
reflected (and less transmitted) for one polarization component, but the
other component exhibits a minimum at a special angle at which no light
is reflected.
We consider the origin of this phenomena.
When radiation is incident on a material, the oscillating electric field from
the EM wave induces the electrons associated with the atoms near the
surface of the material to oscillate about their nuclei at the same frequency.
For polarized incident radiation, the induced oscillations in the electronic
distribution around the nuclei induce the atom to emit in a so called dipole
radiation pattern.
Colors here indicate
weak
zero
source of light. Out
going radiation has
in
same f (color) as
incoming.
Strong
+
in
slice through
vertical plane
including atom
zero
weak
top view
zero
weak
cylindrical
symmetry
in
Strong
+
in
slice through
vertical plane
including atom
zero
weak
top view
Note that no radiation appears in the direction along the line of the induced
oscillation. The oscillating charge has no component of acceleration
(generally what causes radiation) perpendicular to that direction to launch
EM waves in that direction.
In a transparent material part of this radiation is propagated forward into the
material as the transmitted wave and a fraction is sent out from the surface
as the reflected wave.
The directional nature of the transmitted and reflected rays, that makes them
consistent with Snell’s law and the law of reflection occurs because of
interference between the waves from nearby atoms causing reinforcement
and cancellation of field amplitudes in the various directions (HRW
chapt. 35).
Consider what happens when the incident radiation is polarized in the
plane of incidence ( the plane that includes the surface normal and incident
ray, called p-polarization) along with the condition that θt + θr = 90o.
Requires θt ⊥ θr
p-polarized
θi
θr
+
θt
In that case the oscillating dipoles are ⊥ to θt and along the line of θr , a
direction in which they can not radiate, so all the incident beam is
transmitted and none reflected.
θt + θr = 90o
θi θr
+
θt
This special angle is called Brewster’s angle, designated θB. Since the
law of reflection requires that θr = θi we also have that, θi = θB . Now,
θB + θt = 90o
θt = 90o– θB
Snell’s law then gives (with θi = θB),
n1 sin θB = n 2 sin θt = n 2 sin(90o − θB ) = n 2 cos θB
n1 sin θB = n 2 cos θB
sin θB n 2
=
cos θB n1
n2
tan θB =
n1
⎛ n2 ⎞
θB = tan ⎜ ⎟
⎝ n1 ⎠
−1
If medium 1 is air (n1~ 1)
θB = tan −1 n 2
If medium 2 is glass of index n2 = 1.5, θB = 56.3o.
Applications
θB
vacuum
When high power lasers must be
coupled into vacuum systems through
windows (to maintain the vacuum),
Brewster angle windows are used to
avoid the loss of power to reflection.
Note that if the angle of incidence for the p-polarized light goes steeper
than Brewster’s angle the reflected intensity increases again, so Brewster’s
angle constitutes a minimum (= 0) in the reflection.
Light polarized perpendicular to the plane of incidence (called s-polarized
light) induces electronic oscillation parallel to the surface and doesn’t
produce such a minimum.
s-polarized
Unpolarized incident light can be resolved into s and p polarized
components. At Brewster’s angle only the s-polarized component
is reflected so the surface polarizes the light on reflection.
Sunlight reflecting off the surface of water near Brewster’s angle is
horizontally polarized. This is why polarizing sunglasses have their
polarizing axis vertical. To block the polarized s-component that
remains in the refection.
p
s
θB
polarizing direction
vertical (allowed to pass)
blocking horizontal
polarization.
Image formation in mirrors (chapt. 34)
Plane mirrors
p
object
Rays of scattered light
diverge from each
point of the object.
i
virtual image
same orientation
and size as object.
image is erect
(as opposed to
inverted)
The rays that reflect from the mirror, in the direction of the pupil, extended
back, appear to converge from a virtual image at a distance behind the
mirror that equals the object distance in front of the mirror, |i| = p.
Distances for virtual images are by convention negative so,
i=–p
Spherical mirrors
Two general types:
Concave (Innie)
Convex (Outtie)
The concave mirror
Has radius of curvature r.
Light in and out
Parallel rays of light (plane waves)
coming in along the central axis
reflect to a single point, a distance f
from c, called the real focus, F.
This focal distance is completely
determined by radius of curvature r.
Like a plane mirror the concave mirror also forms images, but
the location, size and type is distinct from the plane mirror.
(CA)
Image location for the concave mirror by ray tracing.
From tip of object draw two rays. One
parallel to CA (reflection goes
through focus) and one through focus
(reflection goes parallel to CA).
p
F
Where those rays converge
locates the tip of the image.
CA
i
In contrast to the plane mirror, the image here
is a real image (a replica of the object will
exist on a screen placed in the image plane).
f
p
Note also that the image is inverted and
a different size than the object.
CA
As the object moves toward the mirror,
the image moves away from the mirror
and grows (magnifies).
F
i
If the object is closer to the concave mirror
than the mirror focal length we do not get
a real image.
Again consider 2 rays coming from the
object tip.
The ray along the direction from the focus to
the mirror (green ray), must reflect back
parallel to the CA.
p
CA
F
f
i
The other ray parallel to the CA (black ray),
must, on reflection, go through the focus.
Note that these two reflected rays diverge so there is no real image,
however, their projections behind the mirror surface converge to form
a magnified, erect, virtual image, that an observer sees.
The convex mirror
Also has radius of curvature r.
Light in and out
Parallel rays of light (plane waves)
coming in along the central axis
diverge as if coming from a single
point, a distance f from c, called the
virtual focus, F.
This focal distance is completely
determined by radius of curvature r.
Also forms (virtual) images.
(CA)
Image location for the convex mirror by ray tracing.
Again draw two rays from the tip of the
object.
Draw one ray parallel to the CA
(green ray), which must reflect
along the line to the virtual focus.
f
p
CA
F
Draw the other ray (black) toward the virtual
focus, which must reflect parallel to the CA.
i
Since these reflected rays diverge they can not form a real image.
Their projections back into the mirror, however converge, so we
again get a virtual image, which in this case is erect and smaller.
Note that as the object approaches the mirror, the black ray from the tip
going toward the focus steepens raising the reflection (parallel to CA)
making the image approach the surface and grow. You’ve all probably
experienced your nose growing as you approach a convex mirror.
These geometrical constructs are useful for understanding the image
formation characteristics of these mirrors, and very carefully made drawings
and measurements can give quantitative information but an algebraic
formulation is preferred.
Two simple relations (also useful for image formation with thin lenses) and
some conventions (rules) provide this (HRW sec. 34-9 for proof).
1 1 1
+ =
p i f
Where, as above (all measured along the central axis),
p = object-mirror distance
i = image-mirror distance
f = focal length (real or virtual)
And,
i
m=−
p
Where m is the lateral magnification (i.e. image height/object height).
1 1 1
+ =
p i f
,
m=−
i
p
The conventions are as follow:
The object distance p is positive.
The focal length f is positive for a real focus (concave mirror) and
negative for a virtual focus (convex mirror).
The image distance i is positive for a real image (the image in front of a
mirror) and negative for a virtual image (the image behind the mirror)
A positive magnification, m, means that the object and image have the
same orientation while negative means they have opposite orientation.
Example
A 4 cm high toy car lies 25 cm (along the CA) from a concave mirror
that has a 50 cm focal length.
Where is the image?
1 1 1
+ =
p i f
1 1 1
+ =
25 i 50
positive because concave
real focus
1 1
1
1
2
1
=
−
=
−
=−
i 50 25 50 50
50
i = −50cm
Negative so virtual image, 50 cm behind mirror.
What’s the magnification and orientation?
m=−
−50
i
=−
=2
p
25
Image is twice as large as object and erect
(positive sign means erect or same orientation).
Example
A 4 cm high toy car lies 25 cm (along the CA) from a convex mirror
that has a 50 cm focal length.
Where is the image?
1 1 1
+ =
p i f
1 1
1
+ =
25 i −50
negative because convex
virtual focus
1
1
1
1
2
3
=− −
=− −
=−
i
50 25
50 50
50
i = −16.7cm
Negative so virtual image, 17 cm behind mirror.
What’s the magnification and orientation?
m=−
−16.7
i
=−
= 0.67 Image is 2/3 as large as object and erect.
p
25