Tutor. A B (Plate 11. Fig. 16.) represents a concave mirror, and a b, c d, ef, three parallel rays of light falling upon it. c is the centre of concavity, that is, one leg of your compasses being placed on c, and ther open them to the length c d, and the other leg will touch the mirror A B in all its parts. James. Then all the lines drawn from c to the glass will be equal to one another, as c b, c d, and cƒ? Tutor. They will: and there is another property belonging to them; they are all perpendicular to the glass in the parts where they touch. Charles. That is c b, and c fare perpendicular to the glass at b and ƒ, as well as c dat d. Tutor. Yes, they are:-c d is an incident ray, but as it passes through the centre of concavity, it will be reflected back in the same line, that is, as it makes no angle of incidence, so there will be no angle of reflection: a b is an incident ray, and I want to know what will be the direction of the reflected ray? Charles. Since c b is perpendicular to the glass at b, the angle of incidence is a b c; and as the angle of reflection is always equal to the angle of incidence, I must make another angle, as c b m, equal to a b c,* and then the line bm is that in which the incident ray will move after reflection. Tutor. Can you, James, tell me how to find the line in which the incident ray ef will move after reflection? James. Yes: I will make the angle cfm equal to cfe, and the line fm will be that in which the reflected ray will move; therefore efis reflected to the same point m as a b was. Tutor. If, instead of two incident rays, any number were drawn parallel to cd, they would every one be reflected to the same point m; and that point which is called the * To make an angle c 6 m, equal to another given one, as a b c. From 6 as a centre with any radius ↳ x describe the arc xo, which will cut e b in z, take the distance z in your compasses, and set off with it z o, and then draw the line bo m, and the angle mc is equal to the angle a b c, 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 AB (Plate II. Fig. 17.) be, a concave mirror, and M 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 s 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. Charles. Then if I draw c A, that line will be perpendicular to the glass at the point a : the angle M A C is now given, and it is the an gle of incidence. James. And you must make another equal to it as you did before. Tutor. Very well: make c A x equal to MAC, and extend the line A x to any length you please. Now you have an angle м 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 z equal to it, and the line c z being produced, cuts the line a x in a particular point, which I will call m. Tutor. Draw now the perpendicular C B, and you have with it, and the ray м B, the angle of incidence м BC: make another angle equal to it, as its angle of reflection. James. There it is C B u, and I find the line B u meets the other lines at the point m. Tutor. Then m is the point in which all the reflected rays of м will converge; of |