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SPH3UW: OPTICS II
h
s’
s
f
h’
Introduction to Lenses
Refraction
Light passes through all transparent media as well as vacuum.
When light enters from one medium to another, it suffers a
deviation from its path, caused by the difference in its speed.
This phenomenon is known as REFRACTION
Incident ray
Medium 1
Medium 2
LAWS OF REFRACTION
First law: The incident ray, the refracted ray and the
normal to the refracting surface at the point of
incidence are in the same plane.
i
normal
Medium 1
Incident ray
r
i – angle of incidence
Medium 2
Refracted ray
r – angle of refraction
Second law – Snell’s law: The sine of the angle of
incidence is directly proportional to the sine of the
angle of refraction.
Snell’s law :
sin 1  n2

sin  2  n1
Refractive Index
n1 sin 1   n2 sin 2 
Sometimes
written this way
normal
Incident ray
n1
n2
θ1
Medium 1
Medium 2
θ2
Refracted ray
n is the Refractive index of a medium. It is the ratio of the
velocity of Light in a vacuum to the velocity of the speed of
c
light in the medium.
n
v
Refractive index is a quantity with no unit.
Snell’s Law Practice Question
Light strikes a flat piece of glass at an incident angle of 600. If the
index of refraction of the glass is 1.5.
a) What is the angle of refraction when the light enters the glass (r)?
b) Exits the glass back into the air (e)?
Snell’s Law Practice Question
Light strikes a flat piece of glass at an incident angle of 600. If the
index of refraction of the glass is 1.5.
a) What is the angle of refraction when the light enters the glass (r)?
b) Exits the glass back into the air (e)?
Apply Snell’s Law
ni sin i   nr sin r 
ni
sin  r   sin i 
nr
1.00
sin  60 
1.50
 0.57735
 r  35.3

Snell’s Law Practice Question
Light strikes a flat piece of glass at an incident angle of 600. If the
index of refraction of the glass is 1.5.
a) What is the angle of refraction when the light enters the glass (r)?
b) Exits the glass back into the air (e)?
Apply Snell’s Law
ni sin i   nr sin r 
ne
sin  r   sin  e 
nr
1.50
sin  35.3 
1.00
 0.86603
 e  60

Note: that the direction of the beam remains unchanged after
passing through a pain of glass.
Total Internal Reflection

At critical angle, refraction no longer occurs
 thereafter, you get total internal reflection
 for glass, the critical internal angle is 42°
 for water, it’s 49°
 a ray within the higher index medium cannot escape at shallower angles
(look at sky from underwater…)
Critical Angle
AIR
WATER
Total
Internal
Reflection
Light
Source
Understanding Refraction

Think of refraction as a pair of wheels on an axle
going from sidewalk onto grass
 wheel moves slower in grass, so the direction
changes
Note that the wheels
move faster (bigger space)
on the sidewalk, slower
(closer) in the grass
Spring 2006
UCSD: Physics 8; 2006
9
Understanding Total Internal
Refraction
Wheel that hits sidewalk starts to go faster,
which turns the axle, until the upper wheel
re-enters the grass and goes straight again
Spring 2006
UCSD: Physics 8; 2006
10
Critical Angle
Whenever light goes from a medium with a lower index of refraction to one with
a higher index of refraction, the light ray is bent towards the normal. If the ray
goes from a medium with a higher index of refraction to one with a lower index of
refraction, the light ray is bent away from the normal. The angle is determined by
Snell’s Law
ni sin i   nr sin r 
What happens if θr is 900?
ni sin i   nr sin  90 
ni sin i   nr
sin i  
nr
ni
sin c  
nr
ni
The light ray then does not pass through the boundary, but when the angle of
incidence is greater than this critical angle, the ray undergoes a reflection (total
internal reflection). In this case Snell’s Law tells us that the sine of the angle of
reflection is greater than unity, which is impossible
Lenses

A lens is a piece of transparent material shaped such that parallel light
rays are refracted towards a point, a focus:

Convergent Lens
Positive f
 light
moving from air into glass
 will move toward the normal
 light moving from glass back into
 air will move away from the normal
 real focus
– Divergent Lens
 light moving from air into glass
 will move toward the normal
 light moving from glass back into
 air will move away from the normal
 virtual focus
Negative f
Thin Lenses


A thin lens consists of a piece of glass or plastic,
ground so that each of its two refracting surfaces is a
segment of either a sphere or a plane
Lenses are commonly used to form images by
refraction in optical instruments (cameras, telescopes,
etc.)
Converging Lens
Diverging Lens
Focal Length of Lenses

The focal length, ƒ, is the image distance that
corresponds to an infinite object distance
 This

is the same as for mirrors
A thin lens has two focal points,
corresponding to parallel rays from the left
and from the right
A
thin lens is one in which the thickness of the
lens is negligible in comparison with the focal
length
Focal Length of a Converging Lens



The parallel rays pass through the lens and converge at the
focal point F
The parallel rays can come from the left or right of the lens
F is positive
Focal Length of a Diverging Lens



The parallel rays diverge after passing through the diverging
lens
The focal point is the point where the rays appear to have
originated
f is negative
The Lens Equation

We now derive the lens equation which determines the image distance in terms of the object
distance and the focal length.

Convergent Lens:
h
s’
s
f
f
Ray through focus will
refract parallel
h’
Ray Trace:
• Ray through the center of the lens (light blue) passes through undeflected.
• Ray parallel to axis (white) passes through focal point f.
These are principal
rays!!!
h
s
h
h
two sets of similar triangles:
h
eliminating h’/h:

s
s s  f

s
f
magnification: also same as mirror eqn!!
M < 0 for inverted image.
s  f
1 1 1
 
s s f

f
same as mirror eqn
if we define
s’ > 0
f>0
s
M 
s
Lens Equation

The equation can be used for both converging
and diverging lenses
 A converging lens has a positive focal length
 A diverging lens has a negative focal length
1 1 1
 
s s f
Sign Conventions for Thin Lenses
Quantity
Positive When Negative When
Object location (s)
Object is in front of
the lens
Object is in back of
the lens
Image location (s’ )
Image is in back of
the lens
Image is in front of
the lens
Image height (h ’)
Image is upright
Image is inverted
R1 and R2
Center of curvature
The radii of curvature is in back of the lens
Center of curvature
is in front of the lens
Focal length (f )
Diverging lens
Converging lens
Image Formation in a Lens
Ray tracing Technique:
1. A ray which leaves the object parallel to the axis, is refracted to pass
through the focal point.
2. A ray which passes through the lens’s center is un-deflected.
3. A ray passing through the focal point (on the object side) is
refracted to end up parallel to the axis.
Imaging with a Convex Lens
is bent upon entering the lens. and passes through a point
called the focal point.
Arrow as
Upon exiting the lens it is bent again
Object A ray parallel to the principal axis
sees an
image here.
f
Principal Axis
An eye placed here
Convex Lens
A ray passing through the center of the lens is basically undeflected.
This arrangement produces an inverted, real, diminished image.
More Imaging
With a Convex Lens
is bent upon entering the lens.
Upon exiting the lens it is bent again
Arrow as
and passes through a point
A ray parallel to
the principal axis
Object
called the focal point.
sees an
image here.
f
Principal Axis
Convex Lens
An eye placed here
A ray passing through the center of the lens is basically undeflected.
Farsighted
people
usemagnified
lenses similar
to these.
This arrangement produces
an upright,
virtual,
image.
It is a simple magnifying glass.
Converging Lenses
Imaging with a Concave Lens
is bent upon entering the lens.
Arrow as
Object A ray parallel to the principal axis
Upon exiting the lens it is bent again
such that is appears to have come
from a point called the focal point.
f
Principal Axis
sees an
image here.
An eye placed here
Concave Lens
A ray passing through the center of the lens is basically undeflected.
Nearsighted
peoplean
use
lenses virtual,
similar to
these. image.
This arrangement
produces
upright,
diminished
Diverging Lens
Double
Convex
Double
Concave
Plano
Convex
Plano
Concave
Convex
Meniscus
Convex lenses are positive converging lenses.
Concave lenses are negative diverging lenses.
Concave
Meniscus
Lens Equation summary
• Convergent lens example
h
s’
s
f
h’
1 1 1
 ' 
s s
f
fs
s 
s f
'
s ' h'
M  
s
h
f
M 
s f
Diverging Lens Example
You are given a thin diverging lens. You find that a beam of parallel rays spreads
out after passing through the lens, as though all the rays came from a point 20.0
cm from the centre of the lens. You want to use this lens to form a erect virtual
image that is 1/3 the height of the object. Where should the object be placed?
f = -20cm
m = +1/3
s=?
s‘ = ?
1
3
 s
m
s
m
s
s  
3
1 1 1
 
s s f
1 1
1


s  s 20.0cm
3
1 3
1
 
s s
20.0cm
s  40.0cm
s
3
40.0cm

3
 13.3cm
s  
The image is negative, so the object and
the image are on the same side of the lens
Lens practice question
A (half covered) lens is used to image
an object on a screen. The right half of
the lens is covered. What is the nature
of the image on the screen?
object
lens
(a) left half of image disappears
(b) right half of image disappears
(c) entire image reduced in intensity
screen
• All rays from the object are brought to a focus at the screen by the lens.
• The covering simply blocks half of the rays.
• Therefore the intensity is reduced but the image is of the entire object!
•( one more thing… In our examples of image formation, we only needed the top
half of the lens to form the image)
s’
h
s
f
h’
Divergent lens example
Because the lens is divergent, f is negative: -|f|
h
h’
s
1 1
1
 ' 
s s
 f
f
 f s
s 
s f
'
s’
s ' h'
M  
s
h
f
M 
s f
The Lensmaker’s Formula



So far, we have treated lenses in terms of their focal lengths.
How do you make a lens with focal length f ?
Start with Snell’s Law. Consider a plano-convex
lens (1 curved side):
Snell’s Law at the convex surface:
nglassn sin   sin
nairasin a
Assuming small angles,  sin    tan     
n  a
The bend-angle b is just given by:
(Nair=1.0)

b
light ray

h
air
N
air
b  a    n    (n  1)
f
h
R
h

R
tan   
h
h
1

  (n  1)   (n  1)
f
f
f
h
The angle  can be written in terms of R, the radius of curvature of the lens :
Putting these last equations together,
1


 (n  1)  (n  1)
f
h
R
a
h
The bend-angle b also defines the focal length f :
b
a

1
1
 ( n  1)
f
R
h  R
Lensmaker’s Formula
The thin lens equation for a lens with two curved sides is then:
 1
1
1 
  n  1 


f
R
R
 1
2 
 1
1 1
1 

  n  1 



s s
R
R
 1
2 
Two arbitrary indices of refraction
Note: for one surface (Planar):
R1 is positive
(because centre
of curvature C1
is on same side
as outgoing light.
R2 is negative
(because C2 is
on side opposite
outgoing light. f
is positive.
R
1  n2  n1  1
1 




f  n1  R1 R2 
Lensmaker’s Formula
 1
1
1 
  n  1 


f
R
R
 1
2 
Sign convention for f:
Positive for converging lens
Negative for diverging lens
Sign convention for R
Positive if the centre of curvature is behind the lens
Negative if the centre of curvature is in front the lens
R1
R2
+
-
Double
concave: -
+
Double
convex:
Lensmaker’s Example
The biconvex lens has an index of refraction of 1.50. The radius of curvature of the
front (left) surface is 10 cm and that the back surface (right) is 15 cm. Determine
the focal length of the lens.
R1 R2
From the sign
convention we find
that R1 is +10 cm
and R2 is -15 cm.
1 1 
1
  n  1   
f
 R1 R2 
1 
 1
 1.50  1 


 10cm 15cm 
f  12cm
Lensmaker’s Example
The biconcave lens has an index of refraction of 1.52. The radius of curvature of
the front (left) surface is 10 cm and that the back surface (right) is 10 cm.
Determine the focal length of the lens.
R1
C1
From the sign
convention we find
that R1 is -10 cm
and R2 is +10 cm.
R2
C2
1 1 
1
  n  1   
f
 R1 R2 
1 
 1
 1.52  1 


 10cm 10cm 
f  9.6cm
(not in notes)


A “meniscus” lens is one
where both sides are curved in
the same direction.
A meniscus lens always has
(a) positive focal length
(b) negative focal length
(c) can be either
•Depending on which radius is larger, the focal length can have
either sign. In general, the action of the lens is determined by
whether the center of the lens is thicker (converging lens) than the
edge or thinner (diverging).
•Why ever use a meniscus lens? Because some of the aberrations
(described later) may be smaller (especially if used together with
other lenses.
Check our understanding

What happens to the focal length of a
lens when it is submerged in water?

In air, the focal length of a glass lens
(n=1.5) is determined to be fair. When the
lens is submersed in water (n=1.33), its
focal length is measured to be fwater. What
is the relation between fair and fwater?
air
water
fair
fwater
air
water
(a) fwater < fair (b) fwater = fair (c) fwater > fair
•The refraction is determined by the difference in the two indices of refraction
− The smaller the difference,
-> the less the bend
-> the longer the focal length
• Think of the case where it is glass “in glass” ! f → infinity !
• Quantitatively, look at the lensmaker’s formula
1
f water
 nglass
1
 nglass
1
1

 1 

 1
 nwater
 R f air  nair
R
Summary of Lenses and Mirrors

We have derived, in the paraxial (and thin lens) approximation, the
same equations for mirrors and lenses:
1
1
1


s
s
f
M 
s
s
when the following sign conventions are used:
Variable
Mirror
Lens
f>0
f<0
concave
convex
converging
diverging
s>0
s<0
real (front)
Same side as light
Opp side as light
s’ > 0
s’ < 0
real (front)
virtual (back)
real (back)
virtual (front)
Principal rays “connect”
object and image
one goes through the
center of the lens or
mirror
the other goes parallel to
the optic axis and then is
refracted or reflected
through a focal point
the third one is like the
second one….
Optical Aberrations
Why really good lenses cost a lot
Chromatic aberration
Due to dispersion (index of refraction
depends on frequency), focal length
can be different for different colors.
Spherical aberration
Outside the “paraxial limit”, optimal
focusing occurs only for a parabolic
lens. Spherical lenses look
~parabolic for narrow field of view.
Astigmatism
Curvature of the lens not symmetric
in transverse directions → slightly
cylindrical → different focal lengths
(But spherical
lenses are much
cheaper to grind
& polish!)