For a prism with a refractive index of 1.5,wh | vedclass

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(C) The formula for the refractive index of a prism is given by $n = \frac{\sin(\frac{A + \delta_m}{2})}{\sin(\frac{A}{2})}$.Given that the angle of m

Vedclass · Accepted Answer

$82$

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(C) The formula for the refractive index of a prism is given by $n = \frac{\sin(\frac{A + \delta_m}{2})}{\sin(\frac{A}{2})}$.Given that the angle of minimum deviation $\delta_m$ is equal to the prism angle $A$,we substitute $\delta_m = A$ 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 $\sin(A) = 2 \sin(\frac{A}{2}) \cos(\frac{A}{2})$,we get:$n = \frac{2 \sin(\frac{A}{2}) \cos(\frac{A}{2})}{\sin(\frac{A}{2})} = 2 \cos(\frac{A}{2})$.Given $n = 1.5$,we have $1.5 = 2 \cos(\frac{A}{2})$,which implies $\cos(\frac{A}{2}) = 0.75$.Given $\cos 41^\circ = 0.75$,we have $\frac{A}{2} = 41^\circ$.Therefore,$A = 82^\circ$.

For a prism with a refractive index of $1.5$,what is the prism angle $A$ if the angle of minimum deviation is equal to the prism angle (in $^\circ$)? (Given: $\cos 41^\circ = 0.75$)

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