Page:Optics.djvu/73

 That is, calling the deviation $$\delta,$$ we have

$$\delta = \phi + \psi - \iota.$$

68. If the angle of the prism, and the angles of incidence and emergence be exceedingly small, we may without great inaccuracy substitute for the sines of $$\phi, \phi', \psi, \psi'$$ the arcs themselves which measure those angles. Then we have

$$\phi = m \phi'; \quad \therefore \phi' = \frac{\phi}{m}, $$

$$\psi = m \psi' $$

$$\psi' = \iota -\frac{\phi}{m}, $$

$$\psi = m \iota - \phi, $$

$$\delta = m \iota - \iota = (m-1) \iota. $$

69. From this equation we may find the value of $$m,$$ if $$\iota$$ be known, and $$\delta $$ observed, for

$$m-1 = \frac{\delta}{\iota}; \quad \therefore m = 1 +\frac{\delta}{\iota} = \frac{\delta + 1}{\iota}. $$

70. The reader may observe, that in the figure we have been using, the ray is, by the refraction, bent away from the angle $$I$$ of the prism; this is universally the case, as we may easily show.

Let us take the three different cases.

(1) When the angle $$IRQ$$ is an obtuse angle, (Fig. 61.)

(2) When it is a right angle, (Fig. 62.)

(3) When it is an acute angle, (Fig. 63.)

In the first case, $$IRS $$ and $$ ISR$$ are both acute angles, and it is clear, that the bending of $$RST $$ is from the perpendicular $$Sy, $$ that is, from the angle $$I.$$

In the second there is no inflexion at $$R,$$ but $$ISK$$ is an acute angle, and therefore the emergent ray is on the far side of the perpendicular from $$I,$$ and of course declining from it.

In the third case, the deviation is at first towards the angle, afterwards from it, and we must show that the second deviation exceeds the first.

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