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Reflection and Refraction
of Light
Amy C. Nau, O.D., F.A.A.O
Objectives
• Index of Refraction
• Laws of Reflection
• Laws of Refraction
Index of Refraction
• Recall that in in media other than a
vacuum, waves travel more slowly and
their wavelength also decreases v=fl
• The ratio of velocity of light in a vacuum
divided by velocity in another medium is
the INDEX OF REFRACTION (n)
• n=velocity in vacuum/velocity in medium =
c/vm=3.0x108m/s/vm
Index of Refraction
• Since c is always the greatest, n is always
greater than 1.
• It is convention to treat the nair as 1.0
–
–
–
–
–
–
–
–
Vacuum 1
Air (nonpolluted) 1
Water 1.33
PMMA 1.49
Crown glass 1.523
Diamond 2.417
Cornea 1.376
Zeiss hi-index 1.8
Index of Refraction
• Clinically, patients ask about “high index”
glasses. This means that the lenses are
thinner, and more light is bent per unit
thickness of glass.
• The velocity of light decreases and
wavelength also decreases, so the nature
of light changes, which can take some
adjustment. The aberrations also can
increase….
Dispersion and Index
• We know velocity of light varies with wavelength
• The index (n) also varies with wavelength
– Depends on the optical material (dispersion curve)
• Higher index = more dispersion
– Index for shorter wavelength is greater than for longer
wavelengths
• Index of blue light( 1.6 at 400 nm) is greater than for red
(1.47 at 400 nm)
So, as wavelength increases, velocity
increases and index of refraction
decreases.
Reflection
Some definitions
• INTERFACE: boundary between two
media with different indices
Water
n-=1.33
Air
N=1
Regular (specular) reflection
• Reflected light leaves the surface in
definite beams that follows the laws of
reflection. Occurs at smooth surfaces.
Think of throwing a ball against a wall, or playing pool
Diffuse Reflection
Incident light is reflected in all directions and the directions cannot be predicted
Occurs at irregular surfaces.
Reflection
Specular Reflection
http://www.glenbrook.k12.il.us/gbssci/phys/Class/refln/u13l1d.html
Reflection at Plane Surface
N
The incident angle determines the final direction of the ball
Incident and emergent angles are measured from a perpendicular
line (normal) to the surface (interface).
Law of Reflection
Incident
ray
Normal
Incident angle
Reflected angle
Reflected
ray
Medium 1 (n)
interface
qi
qr
Medium 2 (n’)
Refracted ray
The angle of incidence is equal to the angle of reflection qi=qr
Problem
• Ray incident upon two surfaces that are at
right angles. Incident angle is 20 degrees.
Calculate the reflected angle at each
interface and trace path of rays.
Fresnel’s Law of Reflection
• The amount of light that is reflected depends on:
–
–
–
–
Color (black absorbs, white reflects)
Surface (smooth v rough)
Angle of incidence (diffuse reflection only)
Refractive index (type of medium)
% Reflected = (n’-n/n’+n)2 X 100
n’=to left of interface and n= to right of interface…….
Problem
• How much light is reflected at an air water
interface? How about water to glass?
%R= {(1.33- 1.0)/1.33+1.0)}2 X 100
Answer 2.01% Hirschberg Reflex
%R= {1.523-1.0/1.523+1.0)}2 X 100
Answer 4.2% Glasses, lenses etc.
% Reflected = (n’-n/n’+n)2 X 100
Refraction of light
Interface
• The boundary between two media with
different indices
n
n’
air
water
n
air
n’
n
n’
glass
glass
water
Snell’s Law (of Refraction)
nsinq = n’sinq’
n= index of material before refraction
n’= index of material after refraction
q = incident angle
Incident angle
q’= refracted angle
Optical inferface
n
n’
Surface normal
Refracted angle
Normal Incidence
• If a ray of light is travelling perpendicular
to the optical interface. The ray of light will
change speed but not direction.
Refraction problem
• Ray strikes air water interface at 30
degrees. What is the angle of refraction?
nsinq = n’sinq’
1.0 sin 30 = 1.33 sin X
Sin X= (1.0)(.5)/1.33 = .3759
X= sin-1(.3759) = 22.08 degrees
When light travels from medium of lower index to higher index, it bends toward the
Surface normal. When traveling from higher to lower, it bends away from normal.
Critical Angle- (n>n’)
qc=sin-1(n’/n)
• For a high-low (water
to air ,etc) interface,
there is a physical
limit of 90 degrees for
the refracted angle
• The incident angle
that yields a refracted
angle of 90 degrees=
CRITICAL ANGLE
Critical Angle Problem
• Find the critical angle at an interface
between glass (n=1.50) and air
qc=sin-1(n’/n)
= sin -1 (1.0/1.5) = .667
= sin-1(.667) = 41.80
Critical Angle and
Total Internal Reflection
N
• Some general rules:
– When light travels from low
index to hi index interface,
refracted ray bends
towards the normal and
refracted angle is smaller
than incident angle
– When light travels from hi
to low, ray bends away
from normal and the
refracted angle is greater
than incident angle
air n
water n’
cornea n
air
n’
Total Internal Reflection
•If incident angle is larger than the critical angle, it is
“undefined” and the refracted ray is reflected back into the
same medium at the same angle.
glass
air
Used to design reflecting prisms in low vision telescopes,
and to calculate n of unknown materials
Why can’t we visualize the TM?
• The incident angle of the slit lamp beam is larger
than the critical angle and the light is reflected
back at you.
The cornea air interface (high n to low n) causes
TIR
The gonio lens has a higher n than the cornea, so
total internal reflection cannot occur. (eliminates
air surface at the cornea)
TIR cannot occur when light travels from
lower to higher index,so light enters the contact
lens and is reflected from the mirror.
Refraction through parallel sided
elements
n2
n1
Internal
angles
n3
Emergent
angle
Incident
angle
Opposite internal angles are equal
If n1=n3 then incident angle= emergent angle
Refraction through parallel sided
elements
Use Snell’s law repeatedly to figure out all angles
1.0sin25=1.5sin?=16.3
1.5sin 16.3=1.33sin?=18.45
air
n1
n2
1.5
25
1.33sin 18.45=1.5sin?=16.3
1.5sin16.3=1.0sin? = 25!!
n3
1.33
air
n4
1.5
n5
Lateral displacement
The perpendicular distance between an
incident and emerging ray after traveling
through parallel sided elements
air
d
This can affect the apparent
Position of objects……..
Glass plate
n=1.50
air
Why do my feet look weird in the
swimming pool……
• When an object in one medium is viewed
from another medium, the apparent
position of the object differs from the
actual position because of lateral
displacement
• The refracted angle is different than the
incident angle or vice versa depending on
n/l=n’/l’
the media
Apparent position
• Looking from air into water object seems
farther
Looking from water to air
object seems closer
Remember
• Some general rules:
– When light travels from low
index to hi index interface,
refracted ray bends
towards the normal and
refracted angle is smaller
than incident angle
– When light travels from hi
to low, ray bends away
from normal and the
refracted angle is greater
than incident angle
air n
water n’
cornea n
air
n’
Apparent position problems
• Use the formula n/l=n’/l’
• Where n= index where real object is
located
•
n’= index from which it is viewed
•
l= actual distance of object from
interface
•
l’= apparent distance of object from
interface (image)
problem
• A pebble located at the bottom of a fish
tank appears to be 22.5cm from the
surface. What is the depth of the water?
• n/l=n’/l’ so 1.33/l=1.0/22.5 and
• l=(1.33)(22.5)= 30 cm
Fermat’s Principle
• The time for light to
• It can also be
travel a distance (d) in
expressed in terms of
a medium with an
index ( where n=c/v)
index (n) is expressed
t=nd/c
by:
=(c/v)d/c
=cd/cv=d/v
t=dmeters/v sec
The terms nd is optical
path length
Time to travel is nd/c
Fermat’s Principle
• Principle of least time - the path traveled
by light will be that which requires the least
time.
n=1.33
3cm
n=1.55
5cm
n=1.70
10cm
How much time to travel from first to last interface? Just sum the times from
Each interface
T=nd/c=1/c(n1d1+n2d2+n3d3)= 9.58x10-10 sec
Is it moving and shimmering?
Buy a poster!
Look at this illusion for a while and it will
appear to be shimmering and moving.
Also: Follow the outermost groove and watch it
change from a groove to a hump as you go around the wheel.
Curved Surfaces
Object Space
Index before refraction (or reflection) n
Real objects
Virtual images
Image Space
Index post refraction or reflection n’
Real images
Virtual objects
+ direction of light
Negative measurements
Positive measurements
Curved Surfaces
• Real object- object that emits divergent
wavefronts. Radius and vergence are
negative, Real objects must be to the left
of the interface
• Real Image- image formed with
convergent wavefronts, which have
positive values. Located at the right of the
interface and can be formed on a screen
Curved Surfaces
• Virtual object- when convergent wavefront is
obstructed by an obstacle (i.e. refracting
surface) the position to which the wavefront is
headed is the virtual object position. To the right
of the interface
• Virtual image- when divergent wavefront leaves
a refracting surface, the center of curvature of
the wavefront is the virtual image position.
Located to the left of the interface
Curved Surfaces
• Gaussian Imaging Equation: Used to calculate
object position, image position and power.
• A refracting surface changesn
• the vergence of light incident upon it. The
amount of change is equal to the power of the
surface
• L’=L+F
• L’=emerging vergence after refract
• L= incident vergence
• F= power of the surface
Curved Refracting Surfaces
• Any point on a curved surface reflects light
according to Snell’s law.
• The shape of the surface will determine
the size, position, type and quality of the
resulting image
• Spheres are used for Rx lenses, contacts,
IOL’s, instruments
• We will only consider one surface for
simplicity
Curved Surfaces- some concepts
• Spherical surface is defined by the radius,
curvature or power
– Radius- distance from arc to any point
– Curvature- reciprocal of radius
– Center of curvature - equidistant form any
point on surface
– Sag- perpendicular distance from chord to
surface
– Chord- straight line connecting any two points
on any arc
Curved Surfaces
Center of curvature
radius
sag
chord
Contact lens formula for sag depth
r=y2/2s+s/2
Radius-Power Relationship
• F=n’-n/r
• Clinically, this is commonly used to convert
K readings to radius readings.
• This is also the premise of a “lens clock”,
which is used to measure lens power
• Power is + when wavelength converges
and - for diverging wavelengths.
Curved Mirrors
• F=n’-n holds, but since n’=n for mirrors,
the formula can be rewritten as F=-2n/r
• A concave mirror has + power and a
convex mirror has – power!!
Reflection at a Plane Mirror
Incident ray
Reflected ray
Real space
Real object
Real image
Reflecting surface
Virtual space
Virutal object
Virtual image
Reflection at plane mirror
• Law of reflection states that incident and
reflected angles are equal. Opposite
internal angles aremirror
also equal
Virtual, erect image
Real object
l
l’
l=-l’
Reflection Basics
• Rays leaving the surface diverge as though they
came from the image position
• The image is virtual, erect and equal
– An image, although reversed, will have the same size
and orientation as the object
• The image distance is equal in magnitude to the
object distance (but opposite in sign)
– If you stand 1m in front of a mirror, you will see your
image 1 m behind the mirror
Plane mirror- practical problem
Due to cost, you build a 12 foot lane, not a 20’ lane. The patient will sit 10’ from the
Front walll, with floor to eye distance of 4.5’. The bottom of a 2x2 eye chart is placed
6’ above the floor on the wall behind the patient. What is the acutal test distance?
How large must the mirror be so the patient can see the entire chart? At what distance
From the floor must the mirror be placed?
Deviation of Rays at Reflecting
Surface
qr
90-qr
90-qi
Rotation of a Reflecting Surface
If you rotate a reflecting surface, the incident and thus, reflected angles change.
The total rotation causes the doubling of the angle of the reflected ray.
a
n2
i
r
n1
Practical Application
• Measuring eye movements
• Mirror mounted on cl that moves with the
eye
• Light reflected off the mirror is rotated
twice the amount of eye movement which
can be measured on a wall
• Not used much b/c infrared devices are
available.
Two Mirror Deviation
• Deviation of reflected ray depends only on the
angle between the mirrors and not on the angle
of incidence.
• If you rotate the mirrors, the direction of the rays
does not change!
b1+b2+a=180
Deviation = 2(180-a)
b1
b2
a
Multiple Images
• Two mirrors facing each other form
multiple (infinite) images of the object
placed between them