166 Dr Searle, Experiments with a prism of small angle 
the collimator tube is coated with a good ‘ optical black,” so much 
light will be reflected from the sides of the tube that it will be — 
difficult to see the faint secondary image. The difficulty may be 
met by placing the goniometer so that there is a considerable 
distance (80—50 cm.) between its lens and that of the collimator. 
The prism is placed as close as convenient to the lens of the 
goniometer. Under these conditions, when the goniometer arm 
is directed to view the secondary image, the light reflected from 
the sides of the tube will give little trouble. 
$9. Practical example. The following observations were 
made, using the same prism as in § 4. A collimator was used. 
Edge of prism turned towards right. 
Goniometer reading for direct rays, 4°04 cm. 
Reading for primary image, 557 cm. Reading for secondary image, 
14°16 cm. 
Central reading of goniometer scale, 10°00cm. Length of arm, 40 cm. 
The tangents of the three deviations from the central position are 
(10 — 4:04) /40=0-1490 ; (10—5°57)/40=0-1108; (14:16 — 10)/40=0-1040. 
The corresponding angles are, oy the Table in § 2, 
0:1490 —0:0011 =0:1479; 0:1108 —0:0004=0°'1104; 0:1040— 0: 0003 = 0°1037. 
Hence D=01479 — 0:1104=0-0375 radians, 
D'=0'1479+0°1037 ==0°2516 radians. 
When the edge of the prism was turned to the left, similar observations 
gave the values D=0:0372, D’=0'2517 radians. 
The mean values are D=0:03735, D’=0°25165 radians. 
Hence, by (18) and (19), 
ID ID Opies) 
PS DD Os 
i=4 (D'—3D)=$ x0°1396= 0698 radians 
Oc Os 
= 1-535, 
§ 10. Determination of the angle between two nearly perpen- 
dicular nurrors. If two plane mirrors are inclined to each other 
at exactly a right angle, any ray which suffers two reflexions, one 
at_ each of the mirrors, leaves the second mirror parallel to its 
original direction. When the angle between the mirrors 1s not 
exactly 90°, the initial and final rays are no longer parallel. If 
the initial ray is parallel to a principal plane of the mirrors, 
z.e. to a plane which cuts both mirrors at right angles, the angle 
between the two rays is twice the difference between 90° and the 
angle between the mirrors. Let AB, AC (Fig. 11) be a section 
of the mirrors by a principal plane and let the angle BAC be 
ta7+90, so that it exceeds a right angle. Let the ray be PQRS — 
and let BQYP=a. Then, if PQ and SR meet in F, FQR=2a. 
Further, 
ARQ=7-—AQR- CARS a—a—(in+0)=47—-a-ZGO, 
and hence FRQ =2A RQ = 7 — 2a — 20. 
