In regard to the myopes, the case will be exactly the reverse. As the defect of their sight is occasioned by a conformation of the eye which unites the rays too soon, and causes the point where the image of objects moderately distant are painted with distinctness, to fall on this side the retina, they will receive relief from concave glasses interposed between the eye and the object; for these glasses, by causing the rays to diverge, remove to a greater distance the distinct image, according to the third experiment: the distinct image of objects which was before painted on this side the retina, will be painted distinctly on that membrane when a concave glass is employed.

Besides, myopes will discern small objects within the reach of their sight much better than the presbyta, or persons endowed with common sight; for an object placed at a smaller distance from the eye, forms in the bottom of it, a larger image, nearly in the reciprocal ratio of the distance. Thus a myope, who sees distinctly an object placed at the distance of six inches, receives in the bottom of the eye an image three times as large as that painted in the eye of the person who does not see distinctly but at the distance of eighteen inches consequently all the small parts of this object will be magnified in the same proportion, and will become sensible to the myope, while they will escape the observation of the presbytæ. If a myope were in such a state as not to see distinctly but at the distance of half an inch, objects would appear to him sixteen times as large as to persons of ordinary sight, whose boundary of distinct vision is about eight inches: his eye would be an excellent microscope, and he would observe in objects what

persons of ordinary sight cannot discover, without the assistance of that instrument.

PROBLEM V.

To cause an object, whether seen near hand or at a great distance, to appear always of the same size.

THE apparent magnitude of objects, every thing else being alike, is greater according as the image of the object painted on the retina occupies a greater space. But the space occupied by an image on the retina, is nearly proportioned to the angle formed by the extremities of the object, as may be readily seen by inspecting fig. 11; consequently it is on the size of the angle formed by the extreme rays, proceeding from the object, which cross each other in the eye, that the apparent magnitude of the object depends.

This being premised, let A B be the object, which is to be viewed at different distances, and always under the same angle. On AB, as a chord, describe any arc of a circle, as A CDB: from every point of this arc, as A, C, D, B, the object A B will be seen under the same angle, and consequently of the same size; for every one knows that all the angles having AB for their base, and their summits in the segment A CD B, are equal.

The case will be the same with any other arc, as Acd B.

PROBLEM VI.

Two unequal parts of the fame ftraight line being given, whether adjacent or not; to find the point where they will appear equal.

ON AB and BC (fig. 12 pl. 3), and on the same side, construct the two similar isosceles triangles AFB and BGC; then from the centre F, with the radius F B, describe a circle, and from the point G, with the radius G B, describe another circle; intersecting the former in D: the point D will be the place required, where the two lines appear equal.

For, the circular arcs AEBD and Be CD are, by construction, similar; and hence it follows, that the angle ADB is equal to BD C, as the point D is common to both the arcs.

REMARKS.

1st. There are a great many points, such as D, which will answer the problem; and it may be demonstrated, that all these points are in the circumference of a semi-circle, described from the point I as a centre. This centre may be found by drawing, through the summits F and G of the similar triangles A F B and B G C, the line F G, till it meet AC produced, in I.

2d. If the lines A B and B C form an angle, the solution of the problem will be still the same; the two similar arcs described on A B and B C will necessarily intersect each other in some point D, un

less they touch in B; and this point D will, in like manner, give the solution of the problem.

3d. The solution of the problem will be still the same, even if the unequal lines proposed, A B and bC (fig. 13 pl. 4), are not contiguous; only care must be taken that the radii FB and G b, of the two circles, be such, that the circles shall at least touch each other. If A Ba, Bbc, b C=b, and A C=d=a+b+c, that the two circles touch each other, FB must be at least a√

ab+ ac+be+c2

ab

or a✓ab+cd, and Gb=‡ b ✓✓ab+ac+be+c2

a b

a b + c d

u b

ab

or

If these lines be less, the two circles will neither touch nor cut each other. If they be greater, the circles will intersect each other in two points, which will each give a solution of the problem. Let a, for example, be = 3, b=2, and c=1: in this case Gb will be found 2 1'4142, and BF = √2 = 2°1213, when the circles just touch each other*.

4th. In the last place, if we suppose three unequal and contiguous lines, as A B, BC, CD (fig. 14 pl. 4), and if the point from which they shall all appear under the same angle, be required, find, by the first article, the circumference B E F, &c, from every point of which the lines A B and B C appear under the same angle; find also the arc CEG from which BC and CD appear under the same angle; then the point where these two arcs

* A considerable error in the original has been here corrected, both in the algebraical expressions and in the numeral values.

intersect each other will be the point required. But to make these two circles touch each other, the least of the given lines must be between the other two, or they must follow each other in this order, the greatest, the mean, and the least.

If the lines A B, B C, and CD be not contiguous, or in one straight line, the problem becomes too difficult to be admitted into this work. We shall therefore leave it to the ingenuity of such of our readers as have made a more considerable progress in the mathematics.

PROBLEM VII.

If AB be the length of a parterre, situated before an edifice, the front of which is CD, required the point in that front from which the apparent magnitude of the parterre A B will be the greatest (fig. 15 pl. 4).

TAKE the height CE a mean proportional between C B and C A: this height will give the point required. For if a circle be described through the points A, B, E, it will touch the line C E, in consequence of a well-known property of tangents and secants. But it may be readily seen that the angle A E B is greater than any other A e B, the summit of which is in the line CD; for the angle A e B is less than Ag B, which is equal to AEB.

PROBLEM VIII.

A circle on a horizontal plane being given; it is required to find the position of the eye where its image on the perspective plane will be still a circle.

WE here suppose that the reader is acquainted with the fundamental principle of perspective repre