Today’s Lecture
Brewster’s Angle
Mirrors
Reflection and refraction indices - sparkling diamonds
• For a light beam incident upon a boundary between two
transparent media at 90°:
I r Intensity of reflected beam (n 1 − n 2 )
=
=
2
I i Intensity of incident beam (n 1 + n 2 )
2
Ir
• Glass – n1 = 1.5,
= 0.04
Ii
• Diamond – n1 = 2.4, I r = 0.17
Ii
What happens if the light is not coming at the angle of 90°?
What happens if the light is not coming at the angle of 90°?
The light coming at 90° corresponds to the angle of incidence θ = 0.
As a rule, at greater angles of incidence more light gets reflected.
Look at your watch glass at a grazing angle! It will act as a pretty
good mirror!
Is there any dependence on polarization of light?
How is the reflected wave created in the first place?
How does it depend on the medium the wave is going to?
The primary source of the reflected wave are oscillating charges in
the medium the light is going to.
Therefore the direction of polarization corresponds to the direction
of oscillation of the charges.
EM Waves by an Antenna
• Two rods are connected to an ac source, charges oscillate
between the rods (a)
• As oscillations continue, the rods become less charged, the
field near the charges decreases and the field produced
at t = 0 moves away from the rod (b)
• The charges and field reverse (c)
• The oscillations continue (d)
How does that apply to the light waves?
At a particular angle of incidence, called the Brewster angle, the
light polarized in the plane on incidence-reflection-refraction does
not get reflected at all!
You do not need any special coating – just appropriate polarization.
Brewster angle corresponds to the
situation, when the reflected
wave is supposed to go in the
direction of polarization of the
wave inside glass (2nd medium).
The directions of the reflected and
refracted waves are set by laws of
reflection/refraction.
The charges inside glass oscillate
in the direction of polarization (E)
and cannot radiate the reflected
wave.
Brewster Angle
The Brewster angle corresponds to
the situation when the reflected
wave is supposed to go in the
direction of polarization of the
wave inside glass (2nd medium).
The charges inside glass oscillate in
the direction of polarization (E) and
cannot radiate the reflected wave.
Brewster angle can be found from the
condition that the refracted ray propagates at
90° to the reflected ray. Therefore,
angle of incidence + angle of refraction = 90°
Under what conditions
is the polarizing angle
smaller than 45º?
At a particular angle of incidence, called Brewster angle, the light
polarized in the plane on incidence-reflection-refraction does not
get reflected at all!
By the same token, if the incident
light is unpolarized, only the
component perpendicular to the
plane of incidence gets reflected.
Therefore, the reflected light is
perfectly polarized perpendicular
to the plane of incidence-reflectionrefraction
A shop window photographed at the Brewster angle without and
with polarization filter….
Practical conclusions:
If you need to view objects through glass (camera, telescope
etc), try to look at the Brewster angle for air-glass - 56°.
Do not forget a good polarization filter (polarizer)!
Mirror as an image forming device…
Image – a pattern of light that provides visual representation of reality…
Image - the optical counterpart of an object produced by an optical
device as a lens or mirror …
… a likeness of an object produced on a photographic material.
Image formation in a plane mirror.
The image is always there, in a well defined position, whether you
look at it or not.
The image is formed behind the mirror at a distance equal to the
distance from the object, and has size equal to the size of the object.
It is a “virtual” image, since there is no light in the image location.
The light only appears to come from there.
How exactly do we see
(images of the) objects in a
mirror?
Pretty much the same way
as when we are looking at
the objects directly!
And there are always many
more rays emanating from
an object than getting to our
eyes.
That’s how different people
can see the same
object/image
In the overhead view of the figure below, the image of the
stone seen by observer 1 is at C.
Where does observer 2 see the image – at A, at B, at C, at
D, at E, or not at all?
Position of an image is defined just as well as position of the object!
A little geometry demonstrates that its location is independent of
the position of the observer!
Looking into a mirror at yourself:
For an unobstructed, complete view you
only need a mirror, which is a half of
your height.
Question: What is going to happen to your image in the mirror if
you walk away from it?
Answer: Nothing other than it will appear to be further away,
twice the distance of that from you to the mirror.
Mirrors are known to turn left into
right, that is to make the image of
your left hand look as your right
hand.
It is this effect that gives rise to the expression
“A Mirror Image”
The mirrors actually do a very special transformation, known as
inversion, which cannot be reduced to translations and
rotations…
Maybe to turning inside out?...
Chiral objects and
chiral molicules…
Parabolic Mirrors
Properties of a parabolic mirror
Any ray parallel to the mirror axis reflects
through the focal point
A point source of light at the focus will emerge
from the mirror in a beam of parallel rays
They can be used to concentrate light to high intensities – solar energy
They can be used to design light sources, head lights, flashlights, etc.
Parabolic Mirrors Form Images
The image of the pig is so real it seems
that you could reach out and touch it!
Spherical Mirrors
Spherical mirrors are easier to make
Over a small region near the apex
they are the same as a parabolic mirror
Deviation from parabolic causes
“spherical aberration” (more later)
It can be shown, an exercise for the student, that the
distance to the center of curvature is twice the focal length
Spherical Mirrors Form Images
Light passing through the focal point reflects
parallel to the mirror axis and vice versa.
(a) Object beyond C, image is real, reduced, and inverted.
(b) Object between C and F, image is real, magnified, and inverted.
(c) Object inside F, image is virtual, magnified, and upright.
Note – “Real images are always inverted”
Question: Can a parabolic mirror form an image?
Answer: Of course! The above applies, replace C with 2F
Spherical Mirrors Form Images
1. Any ray parallel to mirror axis reflects through the focal point
2. Any ray that passes through the focal point reflects parallel to the axis
3. Any ray that strikes the center of the mirror reflects symmetrically
about the mirror axis
4. Any ray that that passes through the center of curvature returns on
itself
Any two of these rays is sufficient to locate an image
Spherical Mirrors Form Images
Object located at l
Object height is h
Image located at l’
Image height is h’
The shaded triangles are similar triangles
M is the magnification
Image height is negative if
image is inverted
The further from the mirror (object vs image) the larger the object vs image.
Spherical Mirrors Form Images
l
Object located at l
Object height is h
Image located at l’
Image height is h’
The shaded triangles are similar triangles
Mirror Equation
Spherical Mirrors Form Images
Object located at l
Object height is h
Image located at l’
Image height is h’
As it turns out the image may also be virtual. This occurs when object is closer
to the mirror than the focal length. The image is then upright and virtual.
Convex Mirrors Form Virtual Images
l
l’
Object located at l
Object height is h
Image located at l’
Image height is h’
For a convex mirror the focal length is
always negative, − f. From the mirror
equation:
This result implies:
Image is always:
virtual, upright, and reduced
Hubble Mirror
Mirror Equation
A technician stands 3.85m in front
of the Hubble mirror which has a
focal length of 5.52m.
(a) The location of the tech’s image is?
Why is the answer < 0?
The tech is standing inside
the focal length. The image
is located behind the mirror.
Hence the image is virtual!
(b) The magnification of the
tech’s image is?
Hubble Mirror
Mirror Equation
If the technician stands 15m in front
of the Hubble mirror (a) Find the
location of the image and (b) the
magnification of the image.
(a) From the mirror equation:
(a) The magnification is:
The image is real, reduced, and inverted.
Rear View Mirror
Mirror Equation
A rear view mirror shows the image of
a truck located 10m from the mirror.
The focal length of the mirror is −.6m.
(a) Find the position of the image, and
(b) The magnification of the image.
Since the mirror is convex we expect the image to be upright, reduced,
and virtual. (a) From the mirror equation:
(b) The magnification of the image is:
The image is behind the mirror – hence virtual. It is also upright and reduced.
General Properties
From the mirror equation we have
found that the image location is:
The magnification is:
A question one might ask is “Can the magnification ever be 1?”
If the focal length is negative (convex
mirror) then
If the focal length is positive (concave
mirror) then
So it seems that the magnification can never be exactly 1!
General Properties
From the mirror equation we have
found that the image location is:
The magnification is:
Well what about the question, “Can the magnification ever be −1?”
This can only happen for a concave mirror where we find:
So −1 is possible. For this magnification the image and object are both located
at
The expressions above for the image location and magnification should
make clear the summary statements about images from spherical mirrors.
Do you need a mirror to form an image?
Not necessarily. You can do reasonably
well with a flat refracting surface.
• The image formed by a flat refracting
surface is on the same side of the
surface as the object
– The image is virtual
– The image forms between the object
and the surface
– The rays bend away from the normal
since n1 > n2
The image appears closer, I,
appears closer than the
object, O.
Depth Perception Looking into Water
The object is at O, the image is at I
Assume a light ray from the
object has an incident angle of
Assume light ray from the
Image has an incident angle of
If the depth of a pool is p=2m then from
Snell’s law
The depth of the pool appears to be 1.5m
Light bulb demo.