# Calculating angle of min deviation of prism [closed]

Two rays incident with angle 40 and 60 on one face of equilateral triangular prism the angle of deviation are equal .find angle of minimum deviation?

## closed as off-topic by John Rennie, AccidentalFourierTransform, CuriousOne, Gert, user36790 May 14 '16 at 3:49

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The graph of angle of deviation vs angle of incidence is a $U$ shape. The fact that the angle of deviation is the same for these two rays means that a ray which is incident at $40$ degrees to the normal will emerge at $60$ degrees to the normal. This allows us to find the refractive index.

$$\sin i_1 = \sin 40 = n\sin r_1$$ $$n\sin i_2 = \sin r_2 = \sin 60$$

From geometry, $r_1+i_2 = A = 60$ therefore
$$\sin r_1 = \sin (60-i_2) = \sin60\cos i_2 - \cos 60\sin i_2$$ hence
$$\begin{eqnarray} \sin 40 &=& n\left(\sin 60\cos i_2 - \cos 60\sin i_2\right)\\ &=& \sin 60 \;n\cos i_2 - \cos 60 \;n\sin i_2\\ &=& \sin60 \;n\cos i_2 - \cos 60\sin 60\end{eqnarray}$$
$$n\cos i_2 = \frac{\sin 40}{\sin 60} + \cos 60 = 1.24223$$ $$n^2\cos^2 i_2 = n^2 - (n\sin i_2)^2 = n^2 - (\sin 60)^2 = n^2 - 0.75 = 1.54313$$
$$n^2 = 2.29313$$
$$n = 1.51431$$

When deviation is a minimum then
$$\sin \frac{A+D}2 = n\sin \frac A2 = 1.51431\times\sin 30 = 0.75715$$
$$\frac{A+D}2 = 49.2\; degrees$$
$$D = 98.4 - 60 = 38.4 \;degrees$$