Download GEOMETRIC OPTICS

GEOMETRIC OPTICS
Now that we understand the laws of reflection and refraction we can put them to practical
use by designing optical instruments. We begin with the law of reflection – which tells us that
the angle of incidence equals the angle of refraction.
MIRRORS
We start with the simplest possible mirror – a plane mirror. We need some definitions
which will be used throughout geometric optics. These are best seen with a diagram.
We imagine placing optical elements (mirrors, lenses) in a row, perpendicular to a line which we
call the “optic axis”. We then place on object on that line, again perpendicular to it. The distance
from the object to the element is called the “object distance” and denoted by s. The height of the
object (perpendicular to the optic axis) is called the “object height” and denoted by h. The result
is an image located a distance s’ from the element. That distance is called the “image distance”.
The height of the image (again perpendicular to the optic axis) is the “image height” and denoted
by h’. There is another fundamental quantity, called the focal length, which we will define
shortly. Each of these quantities has a direction associated with it. We therefore need a sign
convention. We choose the following. Imagine the light to be flowing as water in a stream. This
gives us a sense of “upstream” and “downstream”, where upstream is where the light has come
from. Then the sign convention is: s is positive if the object is upstream from the element. For s’
we make a distinction between mirrors and lenses. For mirrors s’ is also positive if the image is
upstream from the element. h is positive if the object reaches from the optic axis to above the
axis (erect). Similarly for h’.
With these definitions and conventions we can now consider plane mirrors. The situation
is shown in the following sketch.
Imagine a light ray emitted from the top of the object running parallel to the optic axis. It will
strike the mirror at right angles. By the law of reflection it will be reflected along the same path
as shown. Now consider a second ray directed to hit the mirror at the optic axis:
We now ask where an observer standing to the left of the mirror will think the rays came from.
That “source” of the light she sees is the image of the object, since to her that is where the object
is. Clearly that point is where the rays cross (they originate together). But there is no point on the
left of the mirror where the rays cross. Rather these projections cross BEHIND (to the right) of
the mirror:
Since h’ = h and the angles are equal, triangles ABC and ADE are congruent. Hence we have s’
= s, and the image is the same distance behind the mirror that the object is in front of it. By our
sign convention this makes s positive and s’ negative. We note that both the object and the image
are erect and therefore both h and h’ are positive.
We now introduce a new concept to distinguish between situations where light actually
comes from the image and those where it merely appears to. The first case we call “real images”
and the second “virtual images”. In our case, there is no light behind the mirror, and hence the
image is virtual.
Summarizing we say that plane mirrors produce virtual images the same distance behind
the mirror that the object is in front. We note also that the image is the same size as the object
and erect.
SPHERICAL MIRRORS
We next consider spherical mirrors. These are bit more complicated because of the shape.
To understand them we proceed as before by tracing the path of the rays emitted from a point on
the object. We begin with a ray parallel to the optic axis.
We first need to find where the ray strikes the sphere. We look in the xy plane since we have
cylindrical symmetry about the optic axis. We know that the equation of the circle is:
 x  R 2  y 2  R 2
for the origin chosen. The equation of the incident ray is:
yh
We find the point of intersection as follows:
 x  R 2  h 2  R 2
x 2  2Rx  h 2  0
1/2
2R  4R 2  4h 2 
x
2
1/2
 R  R2  h2 
Clearly we need the - sign. Hence the point of intersection is:
 R  R 2  h 2 1/2 , h 
Next we need the angle of incidence, θ. We find it as follows:
1/2 
1/2

  R  R   R 2  h 2     R 2  h 2 
 tan  
h
h

1/2
  2
R  h2 
We can now find the equation of the reflected ray.
y  mx  b   x tan  2   b
Note that we have 2θ because the angle of incidence equals the angle of reflection. We can now
solve for b.
tan 2 
Let α = tanθ
sin 2
2sin  cos 

cos 2 cos 2   sin 2 
Then
2h
2
1/2
2
2
2

2
2R2  h2  h
R
h

1


tan 2 



1  2 1  2
h2
R 2  2h 2
1

1  2
R2  h2
1/2
Then
1/2
h
2 R 2  h2 
h
2
2 1/2 
R   R  h    b
 R 2  2h 2 
1/2
2 R 2  h2 
bh
 R 2  2h 2 
1/2
y  x
2 R 2  h2 
 R 2  2h 2 
h
h
2
2 1/2 
R   R  h  
1/2
h
2 R2  h2 
 R 2  2h 2 
h
2
2 1/2 
R   R  h  
We now introduce the final concept we mentioned above. We define the “focal point” to
be the point where rays parallel to the optic axis will cross that axis after striking the element.
The “focal length”, f, is then the distance from the center of the element to the focal point. As a
sign convention we take f positive if the element converges the rays and negative if it diverges
them. In the case of a mirror this means that f will be positive if the focal point is upstream.
We now find the focal point as the point where the reflected ray crosses the axis:
1/2
h
F
2R2  h2 
1/2 
h
R  R 2  h2  

1/2
 R 2  2h 2 
 R 2  2h 2 

 R  R 2  h2 
1/2
1/2
2R2  h2  h
2R2  h2 
 R 2  2h 2 
Obviously this is not independent of h, and thus there is not a unique focal point for a spherical
mirror. However suppose we restrict h to be much less than R. Then:
 2h 2 
1/2
R 1  2 

 1 h2  R
 h2 
h2 
R  2h 2 
h2 


R
 R  R 1 
F
  1  2 1 
  R  R 1 
   0 2 
2

 R2 
 2 R2  2
R 
R  2R 2 
h2 
2 1 

2
 2R 
Hence if h<<R all parallel rays will focus at the same point, and f = R/2.
We now find the image in the same fashion we did for plane mirrors.
Triangles OAB and OCD are similar (all angles are equal). Hence:
AB OB
h
S
S'


  h '  h
 h ' S'
CD OD
S
Also triangle OEG is similar to GDC. Thus:
OG OE
f
h
S




 S' f  h ' S'
GD CD
 fS'  SS' Sf 
1 1 1
 
S f S'
or
1 1 1
 
S S' f
These two equations (with the proper sign conventions)
1 1 1
 
S S' f
h
S

h'
S'
will turn out to apply to all situations.
There is however another configuration to consider with this mirror. Suppose the object is
inside the focal point. Then the picture becomes:
Now the image appears behind the mirror and thus s’ is negative. In this case triangles OAB and
OGH are similar giving:
OA AB
S' h '



OG GH
S
h
as before. Also triangles COD and CAB are similar giving:
CO OD
f
h
S
1 1 1


    
CA AB
f  S' h '
S'
S S' f
as before. Hence the same equations work in this case provided we use our sign convention.
We next consider a convex mirror. Since this will diverge the rays we will have a
negative focal length and the picture below:
Now triangles OAB and OCD are similar, leading to:
OA AB
S
h



S' h '
OC CD
as before. Also triangles ECD and EOH are similar giving:
EO OH
f
h
S


 
EC CD
S'
f  S' h '
  S'f  Sf  SS'  
1 1 1
1 1 1
    
S S' f
S S' f
Thus the same equations also work in this case. For convex mirrors the picture is the same when
the object is placed inside the focal length. Hence we find the same equations for all possible
mirrors. Note that a plane mirror is just a spherical mirror with infinite radius, leading to infinite
f and s = -s’ as we found.
Finally we ask what shape would be required to give a focal length independent of h. We
guess it is a parabola of the form:
x  y 2
We then proceed as before:
1/2
x
y 

 intersection at  h 2 ,h 
We know that the slope of the line perpendicular to a curve y(x) is

1
dy
dx
Hence the angle of incidence is
tan  
1
1/2
 2  x   2h
dy

dx
Then
tan 2 
2 sin  cos 
cos 2   sin 2 
4h
2 2

4h
 tan 2  1  4 h 
4 2h 2  1 4 2h 2  1
1  4 2h 2 
 4h 
y  mx  b  
x  b
 4 2h 2  1 
But
h
4 h
4 2 h 2  1
y 
h 2  b  b  h 
4h
2 2
4 h  1
4 2 h 3
4 2 h 2  1
 x  h 2   h
The focal length is given by y = 0. Thus
0
h 
F 
4h
2 2
4 h  1
 F  h 2   h
4h
2 3
2 3
4 2h 2  1  h  4 h  4 h  1
4h
4h
4
2 2
4 h  1
Independent of h!!
For this reason large astronomical telescopes are made with parabolic mirrors. The
mirrors have to be large in order to catch enough light to see distant objects. On the other hand it
is much easier to grind a spherical mirror. Hence amateur telescopes have spherical mirrors and
small diameters to keep h/R small and avoid fuzzy images.