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:
xR2 y22 R
for the origin chosen. The equation of the incident ray is:
yh
We find the point of intersection as follows:
2 xR h22 R x2Rxh022
1/2 2R 4R22 4h 1/2 xRRh22 2
Clearly we need the - sign. Hence the point of intersection is:
1/2 RR22 h ,h
Next we need the angle of incidence, θ. We find it as follows:
1/2 1/2 RRRh 22 Rh 22
hh tan 1/2 Rh22
We can now find the equation of the reflected ray.
ymxb xtan2 b
Note that we have 2θ because the angle of incidence equals the angle of reflection. We can now solve for b.
sin 2 2sin cos tan 2 cos2 cos22 sin
Let α = tanθ
Then
2 2h 1/2 1/2 2 22Rhh22 22 tan 21 Rh 1122 h 2 R2h 2 1 1a222 R h
Then
1/2 2R22 h h1/2 hRRhb 22 R2h22
1/2 2R22 h h 1/2 bh R R22 h R2h22
1/2 1/2 2Rhh2Rhh22 22 1/2 yx h RRh 22 R2h22 R2h 22
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 2R22 h h1/2 hRRh22 22 R2h22 1/2 FRRhR2h 22 1/2 1/2 2R22 h h 2R 22 h R2h22
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:
2h2 R1 1/2 2 hR2hh22222 1hRh FRR111RR10R h2 RR2RRR22222222 21 2R2
Hence if h< 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 CD OD h ' S' S Also triangle OEG is similar to GDC. Thus: OG OE f h S GD CD S' f h ' S' 111 fS' SS' Sf SfS' or 111 SS'f These two equations (with the proper sign conventions) 111 h S SS'f 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 OC CD S' h ' as before. Also triangles ECD and EOH are similar giving: EO OH f h S EC CD f S' h ' S' 111 111 S'f Sf SS' SS'f SS'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: xy 2 We then proceed as before: 1/2 x y intersection at h2 ,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 11 1 tan dy 2x 1/2 2h dx Then 2sin cos tan 2 cos22 sin 4h 22 4h tan 214 h 4h22 1 4h 22 1 1422 h 4h ymxb xb 4h22 1 But 4h 423 h hhbbh2 4h22 1 4h 22 1 4h yxhh 2 4h22 1 The focal length is given by y = 0. Thus 4h23 0Fhh 2 4h22 1 4h h 22 h422 h 4 23 h 1 F 4h 1 4h 4h 4 4h22 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.