of Achromatic Object- Glasses . 
817 
Example 2. 
It is required to determine the radii of curvature for an ob- 
ject glas^ of 6 feet focus, to be formed of Newcastle plate, whose 
index is 1.515, and of Swiss flint, whose index is 1.671, the dis- 
persive ratio being .61 S. 
1.000 
.618 
.887 
10 
8.87 = focal length of plate. 
.618)8.870(6.31 = focal length of flint. 
Tabular radii for disper- ) 1st surface = 6.7131 
sive ratio .613 j 4th surface — 14.1052 
Tabular index plate 1.524 flint 1.585 
Given index plate 1.515 • flint — 1.671 
— .009 + -686 
Correction (f 1st surface . 
Tab. cor. pi. index =: -f 6.46 for flint index = -f .600 
— .009 -f- .086 
— .05814 -f .0516 
+ .05160 
— .00654 — correction. 
6.7181 
1 st surface 6.70666 = corrected radius. 
Correction of 4>th surface. 
Tab. cor. pi. index — + 111.90 flint index — — 58.32 
— .009 -f .086 
— 1.00764 84992 
— 5.01552 46656 
— 6.02316 = correction. — - 5.01552 
14.1052 
4th surface 8.08204 — corrected radius. 
VOL. XIV. NO. 28. APRIL 1826. 
X 
