Question

A 60°prism is of refractive index 1.5.calculate (a)the angle of minimum deviation. (b)the angle of refraction of the light passing through the prism at minimum deviation.(c)the Angle of incidence at minimum deviation.A=60°, n=1.5, dm=?n= sin½(dm+A) Sin½A

Answer 1

a) $\delta_{min}=2\alpha_1-\phi=2\cdot \sin^{-1}[(n\sin\gamma_1)]-\phi=$

$=2\cdot \sin^{-1}[(1.5\cdot\sin(\phi/2))]-\phi=$

$=2\cdot \sin^{-1}[(1.5\cdot\sin(60/2))]-60=37.18°$ . Answer

b) $\gamma_1=\phi/2=60/2=30°$ . Answer

c) $\alpha_1=\gamma_2=\sin^{-1}[(1.5\cdot\sin(60/2))]=48.59°$ . Answer