OPTICS. 



the plane surface R S. Although we 

 know that the surface of standing water 

 is a curve of the same radius as the 

 globe, yet no skill could discover this 

 curvature, or prove its existence in a 

 square foot of a lake at perfect rest ; and 

 yet this square foot is greater in relation 

 to the radius of the glass than the super- 

 ficial space occupied by a ray of light is 

 in relation to the radius of a common 

 lens. When we wish, therefore, to de- 

 termine the direction of the refracted 

 ray at the point C of any curved surface 

 r C s, we have only to draw a line R C S 

 perpendicular to the radius C Q, touch- 

 ing the curve at C, and proceed in the 

 very same manner as if we were dealing 

 with a plane surface R S. In order to 

 illustrate this, let us begin with a sphere. 

 Refraction through a sphere. Let a 

 ray of light A C (Jig. 6.), fall upon a 



Fig. e. 



sphere of glass L H Z I, at the point C, 

 and parallel to G H O I, the axis of the 

 sphere. Through C draw R S perpen- 

 dicular to O C, the radius of the sphere. 

 This line will touch the sphere at C, and 

 the ray A C will be refracted as if it 

 fell upon the plane surface R S. By the 

 same process which we have already ex- 

 plained for the prism, find the refracted 

 ray C C', and through C' where this ray 

 falls upon the back surface L I Z of the 

 sphere, draw R' S' perpendicular to O C' 

 and touching the sphere at C'. Then, 

 by the same process as before, find the 

 refracted ray C' f. Another ray M N 

 parallel to A C, and falling on the sphere 

 at N, as far from H I, the axis of the 

 sphere, as C is, will obviously be re- 

 fracted to/, because the circumstances 

 of the two rays are exactly the same. 

 Hence, these rays will meet at/, which 

 is called the focus of parallel rays. If 

 \ve continue the lines R S, R' S' till they 

 meet at V, it will be seen that the refrac- 



tion of the ray A C through the sphere 

 is exactly the same as it would have 

 been through a prism R V R. 



If we determine by the preceding 

 method the focus / upon the supposi- 

 tion that the sphere is tabasheer, water, 

 glass and zircon , we shall, by measuring 

 I /, the distance of the focus behind the 

 sphere, obtain the following results, 

 the radius O C of the sphere being sup- 

 posed one inch. 



IndexofRefr.AD 



Tabasheer 1.1145 4 feet inches 



Water 1.3358 1 



Glass 1.500 \ 



Zircon 2.000 



When the index of refraction is greater 

 than two, as in diamond, &c.the point/ 

 falls within the sphere. 



The distance of the focus /from the 

 centre O of any sphere may be found by 

 the following rule : Divide the index of 

 refraction by twice its excess above one, 

 and the quotient is the distance O/ 

 which, in glass, is 1 \ the radius of the 

 sphere. 



Refraction through convex lenses. 

 Light is refracted through a convex lens 

 exactly in the same manner as through a 

 sphere, and the progress of the refracted 

 rays may be found by the method already 

 described for a prism and a sphere. Let 

 L L,yf^. 7, be a double convex lens 



Fig. 7. 



whose axis is R C /, and C its middle 

 point, then it will be found that parallel 

 rays, R L, R L, will be so refracted by 

 the two surfaces as to meet at/, which 

 is called the principal focus of the lens. 

 In like manner it will be found that pa- 

 rallel rays R' L, R'C, R' L, and R"L, 

 R" C, R" L, falling obliquely on the lens, 

 will have their foci at/' and/", at the 

 same distance behind the lens. In these, 

 and all other cases, the rays R C, R' C, 

 R" C which pass through the centre C, 

 will be found to proceed to/,// and/", 

 without changing their direction. The 



