W9_parallel_resonance.eps

398 Week 12: Lenses and Mirrors

reflection) and a ray that goes along the axis and is reflected directly back the way it

came. This is a new kind of image – the rays don’t justappearto come from a point in

space (a point that is really in the dark of your closet or medicine cabinet, back behind the

mirror) as they do with a virtual image, theyreallyreach the eye after passing through a

point in space. You could reach out and put your finger through thepoint in space they

appear to be coming from. We call this kind of image areal image, and we need to

be able to determine whether an image is real (the kind of image that can be projected

on a retina, piece of film, wall, projector screen) or virtual (which cannot be projected

at all, since no light actually passes through the image), so be sure you understand the

distinction and can categorize images you determine from e.g. ray diagrams.

We begin by making an essential approximation. We will later talk aboutaberrationsof

lenses and mirrors – things that prevent rays from a single point on the object from d.

One of the most important ones will besphericalaberration – spheres have this annoying

habit of not focussing parallel rays from an object point far from the axis or rays that

are near the axis but that are not approximately parallel to the axisdown to a single

point in the image. We can’t have that, so we insist that the rays we willdeal with be

paraxial– close to the axis and close to parallel. The former means that we strike the

mirror close enough to its center for us to be able to pretend that the deflection occurs

in a (slightly) curved plane; the latter means that small angle approximations will all

work quite well.

s’

r

s

α β γ

θ

l

P P’

Figure 158: The geometry of forming an image in a concave mirror.

Three important lengths are drawn onto the figure:s,s′, andr, as well as the distance

litself. Note well also the four angles:α,β,γand the angle of incidence/reflectionθ.

Since the angles are all small andlis close to a straight line:

α ≈ l

s

(968)

β =

l

s

(969)

γ ≈ l

s′

(970)

(where the result forβ, note well, is exact becauselreally is the length of a circular arc

that is subtended by the angleβ).

We now play games with the triangles in the picture. We use the followingrule several

times: Consider the triangle withα,θand the angleδ(filled in to figure (159)). We can