# Where did this equation come from ∠I+ ∠E = ∠A+ ∠D?

$\angle I +\angle E=\angle A+\angle D$

Angle of incidence + angle of emergence = angle of prism (Normally $60^\circ$) + angle of deviation.

If their sum is not equal,we made personal error in doing an experiment with prism. Please make sense of this equation.

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The following diagram shows the prism with the incoming and outgoing light rays.

If you follow the incident light ray in, it gets bent by an angle $\theta_1 = i- r_1$. If you follow the light ray where it leaves the glass, it gets bent again by an angle $\theta_2 = e - r_2$, so the total deviation is:

\begin{align} D &= \theta_1 + \theta_2 \\ &= i + e - (r_1 + r_2) \end{align}

For the next step look at the triangle formed by the top of the prism and the light ray, and note that the internal angles must add up to 180°. So:

$$A + (90 - r_1) + (90 - r_2) = 180$$

and a quick rearrangement gives:

$$A = r_1 + r_2$$

Now substitute for $r_1 + r_2$ in our first equation and we get:

$$D = i + e - A$$

or:

$$D + A = i + e$$

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+1, nice answer – OmnipresentAbsence Mar 12 '13 at 11:09
Thanks, though sadly I can't claim this is an original answer as you'll come across it in any introductory optics course. – John Rennie Mar 12 '13 at 11:54