330 



Mr. J. W. Gifford. 



[May 7, 



to an angle of 60°, and if D denote this mean, the refractive index ji 

 is given very approximately by the formula* 



/I = sin J (D + 60°)/sin 30°. 



The method has several advantages. In the first place it is not 

 necessary to measure the angles of the prism with accuracy; then, 

 again, if the prism be suitably placed, light reflected from the outside 

 of the base enters the telescope and is only parallel to the rays of the 

 wave-length under measurement if those rays have passed through the 

 prism parallel to its base. Thus we have two images of the slit, one 

 by refraction the other by reflection, in the field of view in the same 

 direction when the condition for minimum deviation is satisfied. When 

 they overlap in the field of view of the eye-piece, we may, therefore, 

 rest assured that we have minimum deviation for the wave-length 

 under observation. By sliding the prism in a direction at right angles 

 to its base, it is easy to regulate the amount of light thus reflected 

 from the outer side of the base. 



2. Instruments. — A special spectrometer by Hilger with objectives of 

 quartz, the collimator provided with bars carrying the spark apparatus 



* The following investigation by Dr. Glazebrook will show the 

 amount of error introduced by the method : — 



Let D be the deviation and A the angle of the prism ; and let D + S 

 be the deviation corresponding to the angle A + a. Then we have 



. D + A .D + 3 + A + a 



sm — - — sin ~ 



.A ' • A + ol 



sm — sm 



9 9 



whence we find, if a be small, 



. D . D + 2A . D 



sin — a sm sin — Tx 



« 2 a 9 - 2 2 D + A 



6 = a . - T , . + — . t =r t tan — 



• A D + A 4 . 9 A oI) + A 2 



sm — cos sm 2 — cos- — - — 



2 2 2-2 



Now, if «i, a 2 , a 3 be the differences from 60° of the three angles of 

 one of the prisms, each measured in circular measure, D, the deviation 

 which would be observed if the angle were 60°, and D + 8 h D + 5 2 , 

 D + 8 3 , the actually observed deviations, then A is 60° and a.\ + oc 2 -4- a 3 

 is zero. Hence, if Jl be the refractive index as found by the method 

 adopted in the paper, and /x the true refractive index, we have 



sin j (D + A) _ = sin ^{D + A + 1 (81 + 8 2 + S 3 )} m 

 {X siniA ' ^ siniA 



