Chapter 7

Spherical Mirrors

A Thick Mirror

Figure 33.

A thick lens with the second surface covered with a reflecting coating as in Fig. 33 can be treated by the matrix method. The rays on reflection are considered by the reflection transformation above to be proceeding in the positive x direction so that the optical system to the left of the reflecting surface at O' must be "reflected" at O ' as well, as shown in Fig. 33. The final distances in the problem are referred to O '', that is, a positive distance to the right of O '' is actually the same distance to the left of O.

Example 2

A biconvex lens 2 cm thick with a radius of 6 cm is silvered on the second surface. The index of refraction of the lens material is 1.5. Where are the focal planes of this thick mirror?

Solution:


$$[{\bf A}]_{oo''}=
\bmatrix{1&0\cr 1/12&3/2}\bmatrix{1&-2\cr0& 1}
\bmatrix{1&0\cr1/3& 1}\bmatrix{1&-2\cr0& 1}
\bmatrix{1&0\cr1/18&2/3}$$

$$=\bmatrix{5/27&-16/9\cr44/81&5/27}$$

Thus f = 81/44 cm = f '.

OF = (81/44) (5/27) = 15/44 cm and
O''F'= 15/44 cm using equations (23) and (24).

The ray diagram is shown in Fig. 33.

Example 3.

As a further example of the use of the focal planes, suppose an object 2 cm high is placed 15/22 cm in front of the vertex O. Where is the image and how large is it?

Solution:

Since x = 30/44 - 15/44 = 15/44
$$(15/44) x'=(81/44)(81/44)$$

or x'= 9.94 cm, or 9.94 + 15/44= 10.28 cm to the right of O '', which is actually 10.28 cm to the left of O. Since m =- (81/44)/(15/44)= -5.4, the image is 10.8 cm in size and inverted (or m = -x '/f ' = -9.94/1.835 = -5.4).


Exercise Solve the above example to find the focal positions using repeatedly the equations for reflection and refraction at a single surface.

Astronomical Reflecting Telescope

Figure 34.

One simple form of the astronomical reflecting telescope is shown in Fig. 34. A parallel beam of light reflected from the spherical concave mirror is brought to the second focal point, F 'M, of this mirror after further reflection by a small plane mirror. The focal point F ' is also the first focal point FL of the eyepiece lens. The plane mirror is sufficiently small so that it interrupts only a small amount of the light falling on the concave mirror. In another common arrangement, the spherical mirror has a relatively small hole in its centre and a small plane (or convex) mirror redirects the light back to a focus at the centre of the large mirror where the focal image is examined with the eyepiece.

Figure 35.

The matrix method may be used to discuss this telescope as indicated in Fig. 35. Then, if |R| is the absolute value of the radius of curvature of the mirror and fL = f 'L equals the focal length of the lens

$$[{\bf A}]_{oo'}=
\bmatrix{1&0\cr-2/R&1}\bmatrix{1&-(|R|/2+f_L)\cr0& 1}
\bmatrix{1&0\cr1/f_L& 1}$$

or
$$\bmatrix{R/2f_L&-(|R|/2+f_L)\cr0& 2f_L/R}$$

The telescopic system is again indicated by the zero in the lower left hand corner of the matrix and the angular magnification, M = \alpha '/\alpha = R /2fL.

In practice, the large reflecting telescopes are usually used simply as cameras with a photographic plate placed in the focal plane of the large reflector. The surface of the reflector is made parabolic in shape to avoid spherical aberrations. Of course a reflector is free from chromatic aberrations as the laws of reflection are the same for all frequencies of light. The mirror may be made large in diameter in order to attain a high angular resolving power and to increase the light flux per unit area (the illuminance) at the photographic plate.