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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 s, the focus of the convex glass, the diverging rays in the space F f 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 then 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 $, (Plate 11. Fig. 11.) the rays, after passing the lens, will meet at x; and this point x 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 fame of the candle.

James. Why so?

Tutor. Because that is the point where the

rays, if they are not stopped, 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 fame of the candle is inverted.

Fames. How is this explained?

Tutor. Let A B C (Plate 11. Fig. 12.) represent an arrow placed beyond the focus

, 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


which come from A, as A d, A e, and a f, will be refracted by the lens, and meet in A. Those which coine from B, as B d, B e, and B f, will unite in b, and those which come from c, will unite in c.

Charles. I see clearly how the rays from Bare refracted, and unite in b; but it is not

C. The

50 evident with regard to those from the extremities A and c.

Tutor. I admit it; but you must remem. ber 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 E in the centre suffers no refraction, B dis re. fracted into b; and if another ray went from n, as n d, it would be refracted to n, somewhere between 6 and a, and the rays from A must, for the same reason, be refract

ed to a.

James. If the subject A B c is brought nearer to the glass, will the picture be removed to a greater distance ?

Tutor. It will : for then the rays will fall more diverging upon the glass, and cannot be so soon collected into the corresponding points behind it.

Charles. From what you have said, I see that if the object A B C be placed in T, the rays, after refraction, will go out parallel to one another; and if brought nearer to the glass than 7, then they will diverge from one another, so that in neither case an image will be formed behind the lens.

James. To get an image, must the object be beyond the focus F?

Tutor. It must: and the picture will be bigger or less than the object, as its distance from the glass is greater or less than the distance of the object; if A B C (Fig. 12.) be the object, cb A will be the picture ; and if cb A be the object, A B C will be the picture.

Charles. Is there any rule to find the dis. tance of the picture from the glass?

Tutor. If you know the focal distance of the glass, and the distance of the object from the glass, the rule is this :

“Multiply the distance of the focus, by the distance of the object, and divide the product by their difference, the quotient is the distance of the picture.'

James. If the focal distance of the glass be seven inches, and the object be nine inches from the lens, I say, X9 63

=31į inches of course the pic. 2

= ture will be very much larger than the ob


ject. For, as you have said, the picture is as much bigger or, less than the object, as its distance from the glass is greater or less than the distance of the object.

Tutor. If the focus be seven inches, and the object at the distance of seventeen inches, then the distance of the picture will be found

7 x 17 119 thus

= 12 inches nearly. 10 10

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