MEASURING REFRACTION INDICES— POLARIZING MICROSCOPES 205 



slope measured faultlessly. As formerly (§ 2) comparing the shift 

 direction of the fringes with the one obtained from a known object 

 shows whether // > //' or /?' > /;. 



7. DETERMINING THE REFRACTION INDEX FROM PATH-DIFFERENCE 



MEASUREMENTS 



The first thing to do is to interconnect the slope a of the surface 

 waves to the slope of the object. 



Let us revert to the diagrammatic object (Fig. 7.25), shown again 

 in Fig. 7.34. Since a is still assumed to be small (greatly amplified 

 in Fig. 7.34), this means that, provided d be large, n is close to /?'. 

 If e is the thickness common to both media, // and n', the optical path 

 along the radius (2) is u'e and, along the radius (1) located at distance x 

 from B, it is //.vtan + (e— .vtan 0)/?'. The path difference between (1) 

 and (2) is, therefore, (>?' — /7).YtanO. The slope a of the surface wave 

 is derived by merely differentiating .v in relation to the expression 

 {n — u) xi2in . Then: 



a = (/?'-//) tan 0. (7.16) 



Flat-tints method 



Let us substitute a for its value in equations (7.12) and (7.13), 

 when u' > n, then: 



77=;/--^. (7.17) 



^tan0 



Therefore // can be determined provided that the object's slope 

 be known. Reverting to the example in § 5 : the tint is blue {b = 0-664//) 

 at BB' and purple (a = 0-565 /n) at Bbo. Knowing that d = 25 n, 

 tan 61 =0-364 {0 -20°) and //' = 1-527, the expression (7.17) gives 

 n = 1-52. 



As mentioned in § 3, accuracy can scarcely exceed 2 places of 

 decimals. 



Fringe-shift method 



When //' > 77, the expression (7.15) shows: 



Following up the same numerical example: d = 25 /n, tan 6 =0-364, 

 n' = 1-527: in mercury green fight (A =0-546/0 the fringe shift ob- 



