The angle of minimum deviation for a prism of | vedclass

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Question

(C) The formula for the refractive index $n$ of a prism in terms of the angle of prism $A$ and the angle of minimum deviation $\delta_m$ is given by:

Vedclass · Accepted Answer

$82$

Vedclass · Answer

(C) The formula for the refractive index $n$ of a prism in terms of the angle of prism $A$ and the angle of minimum deviation $\delta_m$ is given by: $n = \frac{\sin(\frac{A + \delta_m}{2})}{\sin(\frac{A}{2})}$.Given that the angle of minimum deviation is equal to the angle of the prism,i.e.,$\delta_m = A$.Substituting this into the formula: $n = \frac{\sin(\frac{A + A}{2})}{\sin(\frac{A}{2})} = \frac{\sin(A)}{\sin(\frac{A}{2})}$.Using the trigonometric identity $\sin(2	heta) = 2\sin(	heta)\cos(	heta)$,we can write $\sin(A) = 2\sin(\frac{A}{2})\cos(\frac{A}{2})$.Therefore,$n = \frac{2\sin(\frac{A}{2})\cos(\frac{A}{2})}{\sin(\frac{A}{2})} = 2\cos(\frac{A}{2})$.Given $n = 1.5$,we have $1.5 = 2\cos(\frac{A}{2})$,which implies $\cos(\frac{A}{2}) = \frac{1.5}{2} = 0.75$.Given $\cos(41^o) = 0.75$,we equate the angles: $\frac{A}{2} = 41^o$.Thus,$A = 82^o$.

The angle of minimum deviation for a prism of refractive index $1.5$ is equal to the angle of the prism. The angle of the prism is.......$^o$ (Given: $\cos 41^o = 0.75$)

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