Construction of the Photoplione. 289 



R P, P Q, be drawn making equal angles of incidence and 

 reflexion with the surface of the cone. Then, since P R is 

 parallel to the axis of the cone, the angle S P R=Q PO = ^ 

 and the triangle Q P is isosceles. Hence the exterior angle 

 P Q M = 2<9. If P M be drawn from P perpendicular to the 

 axis, its length, which we may call a, will be that of the half- 

 aperture. 



Fig. 1. 



axis 



Now sinPQM = S in20= " . 



therefore a=PQ sin 20, 



a = l sin 20. 

 Hence if I be constant, a will be a maximum when -=^ =0. 



Now -jn =^cos 20 ; and equating to zero we find 

 cos 20 = 0, 

 20 = 90°, 

 or (9 = 45°. 



In other words, the mirror cone must have an apparent ver- 

 tical angle of 90°, and its development will be a sector of 254° 



= = J cut from a circle whose radius is V2 X Q. 



v — ' 



If the cylinder cell have itself a radius r, then the whole 

 diameter of the linear aperture will be equal to 2 (l + r) : and 

 the cone may be conveniently truncated at a distance along 

 the axis from equal to r, which would leave a circle of 2r 

 diameter just fitting the posterior end of the cylinder. 



It may be remarked that the anterior end of the cylinder 

 will, when it is placed in position, be in the same plane as the 



