Page:Optics.djvu/74

 Supposing the ray to proceed both ways out of the prism at $$R$$ and $$S,$$ the angle of incidence at $$R,\ SER,$$ is less than that at $$S,$$ namely, $$ySR,$$ the latter being the exterior angle of the triangle $$SER,$$ the former an interior and opposite angle.

Now the greater the angle of incidence, the greater is that of deviation, for if $$\phi, \ \phi'$$ be the angles of incidence and refraction, $$\phi - \phi'$$ is the deviation.

Then since $$\sin \phi = m \sin \phi',$$

$$\sin (\phi+\phi') - \sin (\phi - \phi')=2 \cos \phi. \sin \phi'$$

$$= 2 . \cos \phi.\frac{\sin \phi}{m}$$

$$=\frac{1}{m}. \sin 2 \phi;$$

$$\therefore \sin \phi - \phi' = \sin (\phi + \phi')-\frac{1}{m}. \sin 2 \phi. $$

as $$\phi$$ increases, $$\phi'$$ increases also, and the sine of $$\phi + \phi'$$ increases faster than that of the larger angle $$2 \phi,$$ so that the whole of the 2nd member of this equation increases; therefore $$\sin \phi - \phi'$$ must increase also, to maintain the equality, and consequently $$\phi - \phi',$$ the deviation must increase.

In the above case then, the deviation at $$S$$ being greater than that at $$R,$$ the inflexion of the ray is, on the whole, from the angle of the prism.

71. In making experiments with a prism through which a beam of light is made to pass in a plane perpendicular to its axis, it will be found, that if the prism be turned on its axis, the deviation of the emergent ray from the incident, will in some cases increase, in others diminish, so as to have a minimum value. Let us see to what case that value answers.

Adopting the same notation as before, we have

$$\delta= \phi+ \psi - \iota;$$$$\therefore d\delta=d\phi+d\psi, $$

$$\sin \phi=m\sin \phi';$$$$\therefore \cos \phi\ d \phi= m \cos \phi'. d \phi', $$

$$\sin \psi = m \sin \psi';$$$$\therefore \cos \psi. d \psi = m. cos \psi'. d \psi'.$$