This expression is much simpler than that of (73), as there are no terms involving x and y2. The first integration can always be performed: it gives

and if y' and y' are the least and greatest values of y for a given value of x (given by the equation to the aperture in terms of x), the first integral is, between these limits,

-sin (vt-B)

sin

όλ

(2π
λ

2π

"

(2π

{(25 (pix+2y)}-sin (25 (pz+qy")}]

Let the integrals of the terms within the brackets (with respect to x and between the proper limits) be P and Q: then

the coefficients of

81. Ex. Let the aperture be a parallelogram whose sides are 2e and 2f in the direction of x and y.

Here

COS (px + qy"
bλ

- COS

(px − qƒ)} = cos {1 (px+qƒ)}

- 2 sin (px). sin (3),

This expression is maximum when p=0, q=0: so that there is a bright point in the place of the image determined by common Optics. It is 0 when p is any multiple of or when

όλ

2e

is any multiple of versed by rectangular dark bars at equal intervals, the intervals in the direction of the length of the parallelogram being shorter than the others. For a given value of p, the brightness is greatest when q = 0, or when q has one of the values

This shews that the screen is tra

which makes

bλ 2πqf maximum. Thus it appears sin 2πqf όλ

that there will be a brilliant point at the center; a four-rayed cross through the center, the rays being interrupted at intervals; and a series of less bright patches in square arrangement in the angles of the cross: also the distances from the center are greater for the red rays than for the blue. When the parallelogram is narrow, the bright parts in the direction of one side form one of the kinds of spectra described by Fraunhofer.

82. Let the aperture be an equilateral triangle. Take x in the direction of the perpendicular to one side, and let the angle opposite this side be the origin of co-ordinates: let e be the whole length of the perpendicular. Then

Hence

y' x tan 30°: y' = + x tan 30o.

πα

(2πα

P = [cos {27" (p-q tan:
tan 30")}-cos 3(p+q tan

COS

tan 30"}]

30°

the value of which from x=0 to xe is found by putting

e for x. And

the value of which from x = 0 to x = e is

όλ

{3 (p-9 tan 30"}]

όλ

{

(p + q tan 30°)
(p+q

a30"}].

1+cos

The sum of the squares is (omitting the factor)

=

2p2+6q tan 30° (p2-q*tan2 30o)*

(2 пе

COS

Ολ

(p-gtan 30"}

(bx (P + g tan 30")}

Let p r cos 0, q=r sin 0: which is the same as referring M to the central point of the screen by polar co-ordinates. Then observing that tan2 30= }, and restoring the factors

32π*r*.* sin2 0 . sin3 (0 – 60o) . sin2 (0 — 120′′)

— sin (~ — 180o). sin (~ —240°) . cos (3)

sin (0

(0–60°)}

The maximum value it will be found is when r = 0, and is

The value is also considerable when =0, or = 60°, or

120°, or = 180°, or = 240°, or = 300°, when it is

(as will be found by commencing with the first integral formula of (80), and, for the ray, making q=0; and, for the central point, making p also = 0). This points out exactly the star-like form observed by Sir J. Herschel (Encycl. Metrop. Light, Art. 772).

83. Let the aperture be a great number m of equal parallelograms of length 2f and breadth e at equal distances 9. Here y=-f, y" =+f: and the expression to be integrated is

If k be the value of x corresponding to the first side of the first parallelogram, that corresponding to the first side of the (n+1)th parallelogram will be k+n (e+g), and that corresponding to its last side k+n (e + g) + e. The integral therefore for the (n+1)th parallelogram is