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# A ray of light passes through an equilateral prism such that the angle of incidence and the angle of emergence are equal to $\large\frac{3}{4}th$ of the angle of prism. The angle of minimum deviation is

$(a)\;15^{\circ} \\ (b)\;45^{\circ} \\ (c)\;30^{\circ} \\ (d)\;60^{\circ}$

Can you answer this question?

Given $A=60^{\circ}$
$i=i'=\large\frac{3}{4}$ A
$i = i'= 45^{\circ}$
$\therefore i+i'=A+\delta$
or $90^{\circ}= 60^{\circ}+\delta$
$\therefore \delta=30^{\circ}$
Note that $i=i'$ is condition for minimum deviation .
Hence $\delta =30^{\circ}= \delta \;min$
Hence c is the correct answer.

answered Feb 18, 2014 by
edited Jul 23, 2014