And I wish you to remember what I have told you before, that whenever the appearance of a given object is rendered larger and brighter, we always imagine that the object is nearer to us than it really is, or than it appears at other times.

James. If there be nothing to receive the rays (Fig. 8.) at ƒ, would they cross one another and diverge?

Tutor. Certainly, in the same manner as they converged in coming to it; and if another glass G, of the same convexity as D E, be placed in the rays at the same distance from the focus, it will so refract them, that, after going out of it, they will be parallel, and so proceed on in the same manner as they came to the first glass.

Charles. There is, however, this difference; all the rays, except the middle one, have changed sides.

Tutor. You are right, the ray B, which entered at bottom, goes out at the top b; and A, which entered at the top, goes out at the bottom c, and so of the rest.

If a candle be placed at f, the focus of the convex glass, the diverging rays in the space Ff G, will be so refracted by the glass, that after going out of it, they will become parallel again.

James. What will be the effect if the candle be nearer to the glass than the point f?

Tutor. In that case, as if the candle be at g, (Plate 11. Fig. 10.) the rays will diverge after they have

passed through the glass, and the divergency will be greater or less in proportion as the candle is more or less distant from the focus.

Charles. If the candle be placed farther from the lens than the focus f, will the rays meet in a point after they have passed through it?

Tutor. They will thus if the candle be placed at g, (Plate II. Fig. 11.) the rays, after passing the lens, will meet at r; and this point a will be more or less distant from the glass, as the candle is nearer to, or farther from its focus. Where the rays meet, they form an inverted image of the flame of the candle. James. Why so?

Tutor. Because that is the point where the rays, if they are not stop

ped, cross each other: to satisfy you on this head, I will hold in that point a sheet of paper, and you now see that the flame of the candle is inverted.

James. How is this explained? Tutor. Let A B C (Plate 11. Fig. 12.) represent an arrow placed beyond the focus F, of a double convex lens d e f, some rays will flow from every part of the arrow, and fall on the lens; but we shall consider only those which flow from the points A, B, and c. The rays which come from A, as a d, a e, and ▲ ƒ, will be refracted by the lens, and meet in A. Those which come from B, as в d, B e, and Bf, will unite in b, and those which come from c, will unite in c.

Charles.

I see clearly how the rays from в are refracted, and unite in b; but it is not so evident with regard to those from the extremities ▲ and c.

Tutor. I admit it; but you must remember the difficulty consists in this, the rays fall more obliquely on the glass from those points than from the middle, and therefore the refraction is very different. The ray B F in the centre suffers no refraction, Bd is refracted into b and if another ray went from N, as N d, it would be refracted to n, somewhere between b and a, and the rays from A must, for the same reason, be refracted to a.

James. If the object A B C is brought nearer to the glass, will the