of Edinburgh y Session 1882-83. 
133 
Let us examine the conditions needed in order that the light 
become horizontal at a lower level — say at the height of half a 
mile. In this case the index of refraction at B is obtained from 
the equality sin A x 3961.167 = 3960,5 x or 
sin A X 1.000 168 = , 
which has possible solutions within two limits — one when 
= 1.000 000, in which case A would be 88° 57', that is, the 
elevation would be 1°”03'; the other when A is 90°, giving 
= 1.000 168, which index of refraction would be due to a pressure 
of 17 inches of mercury. We have no idea of how such a rarefac- 
tion of the air at the height of only 2600 feet could be brought 
about. 
To come still nearer home, let us propose an altitude of only one- 
tenth part of a mile, or 528 feet. We then find that with A = 90°, 
the index of refraction at the culminating point must be 1.000 269, 
which belongs to air under a pressure of 27 inches of mercury. This 
diminution of density cannot be due to the lessened pressure of the 
air, which, instead of 2*5 inches, would only give *6, or about the 
quarter of the required reduction. In lowering the culminating 
point, we come nearer to the conditions of a flat earth, and there- 
fore to the possibility or probability of a reflected image. Let us 
then inquire into the conditions when the light is close to the 
earth along its whole path, and tangent to the surface at C and 
at A, 
If we put h for the height in miles, and A = 90°, the index of 
refraction is 
3960 
or developing the fraction into series — 
7^0 I 5960“ 39602 i 
Even at the absolute limit 7i = lT67, the second and following 
terms of the series are too small to be of any account in our present 
inquiry, and thus we may hold that the diminution of the index 
would need to be at the rate of = *000 2526 for each mile of 
5260 
altitude. The diminution of the barometric pressure corresponding 
to this would be 25*26 inches of mercury for each mile, or *00478 
