is proportional to the cosine of the inclination of the direction of emission to the normal* to the element of the surface.

For suppose that a bright body is viewed through a tube of small aperture; when the tube is directed so that the element of the bright surface seen is normal to the line of sight, let the area of the element be w. Then when the tube is directed so that the normal to the element of the bright surface seen through the tube makes an angle ✪ with the line of sight, the area of the element will be w sec 0. Let f(0) be the intensity of emission per unit area in a direction making an angle 0 with the normal to the element; then the whole amount of light transmitted to the eye when the element is inclined to the line of sight at an angle 0 is w sec 0.ƒ(0).

But this, by experiment is independent of 0, and therefore f(0) x cos 0.

Let A be the area of an element of the bright surface and let μA denote the intensity of the light emitted in the direction of the normal to the element. Then μ may be called the intrinsic brightness of the element.

5*. If Q be the quantity of light which falls on an area A of illuminated surface, then Q/A is called the mean intensity of illumination of the area. When the area of the element of bright surfaces is indefinitely small, the mean intensity of illumination is called the intensity of illumination within the element.

We shall now find the illumination of a small area A due to an element of any bright surface B. Let O be the centre of the element of the luminous surface and C the centre of the illuminated area A, and let OC = r. Let 0 be the inclination of OC to the normal at O, and & that of OC to the normal at C.

Then if A subtend a solid angle w at 0, the quantity of light it receives will be μB cos 0 w, where u is the intrinsic brightness of the element.

ω,

μ

* The normal at any point of a surface is the straight line through that point perpendicular to the tangent plane to the surface.

We must next find the value of w.

With centre O and radius OC describe a spherical surface cutting the cone, and let A' be the area of the spherical surface enclosed within the cone. Then the solid angle is given

by the equation

w = A'/r2.

The small areas A and A' may be regarded as plane; the inclination of their planes is 4. Also since A'cuts the generators of the cone at right angles, A' is the projection of the area A, and therefore

and therefore the quantity of light received by A from the element B is

This is symmetrical with regard to the two elements, and would therefore represent the quantity of light received by B from the element A, were it of intrinsic brightness μ.

Leto be the solid angle subtended at C by the bright element B, then it may be shewn as before that

then the illumination of A due to the element B is

με cos φ.

6*. The illumination of a small area A due to any finite surface of uniform brightness may now be found.

Take any small element of the bright surface about a centre O, and as before let σ be the solid angle subtended by it at the centre C of the illuminated area. Let be the angle between the line OC and the normal at 0. Then the illumination due to the bright element is

με cos φ.

is σ;

hence

Describe a sphere of unit radius with C as centre; this will cut the small cone subtended at C by the element of area, in a section whose area o cos will be the projection of this small section on the plane of the illuminated area at C, and may be denoted by ; then the illumination due to the small element is μπ

By addition we arrive at the following method of determining the illumination of a small area at C due to any finite bright surface.

From C'draw radii to all points of the boundary of the bright surface as seen from C, forming a conical surface. Let the part of the surface of the sphere of unit radius whose centre is C, intercepted within the cone, be projected on the plane of the area at C. If be the area of the projection, the illumination of the area will be given by the equation

Ι= μπ.

Ex. To find the illumination due to a spherical luminary.

Let a be the semi-vertical angle of the cone whose vertex is at the centre of the area and which envelopes the bright sphere. The curve in which this cone cuts the sphere of unit radius is a circle whose radius is sin a. Hence if be the zenith distance of the luminary, the illumination on a small horizontal area is

Ι== μπ
sin2a cos 0.

7*. Objects appear equally bright at all distances.

The apparent brightness of an object may be measured by the whole quantity of light entering the eye from the object divided by the area of the picture of the object on the retina of the eye.

Let P be any point of the object, p the corresponding point of the picture on the retina; then it will be shown afterwards that the line Pp passes through the fixed point O, the optical centre of the eye.

Let S be the area of a small object and s that of the picture and let OP = R, Op=r. Then S: R2 = s : r2.

Now the quantity of light entering the eye is μSw/R2, where is the area of the aperture of the eye. This may

p

be written usw/r2; hence, dividing by s, we get the intrinsic brightness of the image, which is equal to μw/r2.

We shall assume for the present that as the eye adjusts itself to different distances, r does not change.

Thus the apparent brightness is independent of the distance of the object.

The area of the aperture of the eye changes according to the brightness of the light. But if we suppose the aperture to remain the same, as the object is removed, no change in brightness has taken place, so that the aperture does not need further adjustment.

When the object is very distant the area of the picture on the eye gets to be very small indeed, so that the nerves of the retina cannot distinguish it from a point. In this case the brightness must be measured simply by the quantity of light; and therefore, by the same investigation, the brightness varies inversely as R2.

EXAMPLES*.

1. The light from two sources is allowed to fall upon the same screen. One light is at a distance a and the light falls directly from it on the screen. From the other, which is at a distance 3a, the light falls at an obliquity of 60°. The illuminations of the screen from the two sources being equal, show that one source is 18 times as bright as the other.

2. A small white surface being placed horizontally on a table, and illuminated by a lamp or candle placed at a given horizontal distance a, show that the height of the flame from the table which will give the greatest possible illumination is equal to a/√2.

3. If candles of equal brightness be placed at the angular points of a regular polygon, prove that a small plane area placed at the centre of the polygon will be equally bright on both sides, whatever be the orientation of its plane.

4. A small plane area is placed at right angles to the axis of a paraboloid of revolution whose convex surface is uniformly luminous. Prove that the illumination produced at the point of the plane where it meets the axis varies inversely as the distance of this point from the focus of the paraboloid.

5. A small plane area is placed parallel to a plane lamina of intrinsic brightness I, of breadth 2a, and of infinite length, at a distance c from the centre of the lamina in a line perpendicular to the lamina. Prove that the illumination at the centre of the plane area is πal(a2+c2).

6. The sides of a triangle are the bases of three infinite rectangles of the same brightness, whose planes are perpendicular to the plane of the triangle; show that all points within the triangle are equally illuminated. Find the position of a point in the plane of the triangle, such that the illuminations at that point received from the three rectangles may be equal.

7. A luminous point is placed on the axis of a truncated conical shell; prove that the whole illumination of the surface of the shell varies as

where a, a' are the radii of the circular ends of the shell and C, the distances of the luminous point from their plane.

8. A right cone of vertical angle 20 is described about a given self-luminous sphere, and at the points of the sphere in which the axis of the cone cuts it, tangent planes are drawn; prove that the mean illumination of that part of the cone which is enclosed between these two planes varies as cos 0 cos2 0.