lel to c d, they would every one be reflected to the same point m; and that point which is called the focus of parallel rays is distant from the mirror equal to half the radius c d.

James. Then we may easily find the point without the trouble of drawing the angles merely by dividing the radius of concavity into two equal parts.

Tutor. You may.-The rays, as we have already observed, which proceed from any point of a celestial object may be esteemed parallel at the earth, and therefore the image of that point will be formed at m.

Charles. Do you mean that all the rays flowing from a point of a star, and falling upon such a mirror, will be reflected to the point m, where the image of the star will appear?

Tutor. I do, if there be any thing at the point m to receive the image. James. Will not the same rule hold with regard to terrestrial objects?

Tutor. No: for the rays which proceed from any terrestrial object, however remote, cannot be esteemed strictly parallel, they therefore come diverging; and will not be converged to a single point, at the distance of half the radius of the mirror's concavity from the reflecting surface; but in separate points, at a little greater distance from the mirror than half the radius.

Charles. Can you explain this by a figure?

Tutor. I will endeavour to do so. Let A B (Plate 11. Fig. 17.) be a concave mirror, and м E any remote

object, from every part of which rays will proceed to every point of the mirror; that is, from the point м rays will flow to every point of the mirror, and so they will from E, and from every point between these extremities. Let us see where the rays that proceed from м to A, C, and B will be reflected, or, in other words, where the image of the point м will be formed.

James. Will all the rays that proceed from м, to different parts of the glass, be reflected to a single point?

Tutor. Yes, they will, and the difficulty is to find that point: I will take only three rays to prevent confusion, viz. M A, M C, M B; and c is the centre of concavity of the glass.

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Charles. Then if I draw c A, that line will be perpendicular to the glass

at the point A: the angle м A C is now given, and it is the angle of incidence.

James. And you must make another equal to it as you did before. Tutor. Very well: make c A r equal to м A C, and extend the line A a to any length you please.

Now you have an angle M C C made with the ray м c, and the perpendicular c c, which is another angle of incidence.

Charles. I will make the angle of reflection c c 2 equal to it, and the line c z being produced, cuts the line A a in a particular point, which I will call m.

Tutor. Draw now the perpendicular c, and you have with it, and the ray м B, the angle of incidence M B C make another angle equal to it, as its angle of reflection.

James. There it is C B u, and I find the line в u meets the other lines at the point m.

M

Tutor. Then m is the point in which all the reflected rays of м will converge; of course the image of the extremity м of the arrow E M will be formed at m. Now the same might be shown of every other part of the object м E, the image of which will be represented by e m, which you see is at a greater distance from the glass than half c c, or radius.

Charles. The image is inverted also, and less than the object.