( <88 ) 
* *1 .1 
A B be a Ray of Light filling on the Speculum at 
B, as is before exprefled, and let it be there re- 
flected towards the Point C of the Lens C G, 
where it is refradted towards the Point D of 
the Lens D F, and there again refradted into the 
Line DE, cutting the Axis in E. The Angle AOP 
contained between this lad Line D E, continued 
backwards, and the firft Line of Incidence of the 
Ray A B, will be very nearly equal to double 
the Angle of Inclination of the Axis of the Lens’s 
E L to the Plane of the Speculum B N ; i. e. dou- 
ble the Angle GHN. 
Demon s t ration. 
Produce the Lines of Incidence and Refledfion 
of the Ray A B and B C, ’till they meet the Axis 
of the two Lens’s in I and L ; and through the 
Point B draw B K perpendicular to the Plane of 
the Speculum, and cutting the fame Axis in K, 
the Angles KBL and K B I are equal. The An- 
gle K LB is the Difference of the Angles I K Band 
KBL; and the Angle H I B is the Sum of the 
Angles I K B and K B I (equal to K B L) : There- 
fore the Angle I K B is equal to half the Sum of 
the Angles H I B and K L B. But by the ’forego- 
ing Lemma, the Angles K L B and FED are very 
nearly equal. Therefore the Angle 1 K B is nearly 
equal to half the Sum of the Angles H I B and 
FED; that is, to half the Angle P O B, and its 
Complement B H I or G H N is nearly equal to half 
the Angle AGP the Complement of P O B to a 
Semicircle. <£. E. T>. 
If 
