The angle of minimum deviation for a prism of | vedclass

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(D) The refractive index $n$ of a prism is given by the formula: $n = \frac{\sin((A + D_m)/2)}{\sin(A/2)}$.Given that the angle of minimum deviation $

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$82.8^{\circ}$

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(D) The refractive index $n$ of a prism is given by the formula: $n = \frac{\sin((A + D_m)/2)}{\sin(A/2)}$.Given that the angle of minimum deviation $D_m$ is equal to the angle of the prism $A$,we substitute $D_m = A$ into the formula:$1.5 = \frac{\sin((A + A)/2)}{\sin(A/2)}$$1.5 = \frac{\sin(A)}{\sin(A/2)}$Using the trigonometric identity $\sin(A) = 2 \sin(A/2) \cos(A/2)$,we get:$1.5 = \frac{2 \sin(A/2) \cos(A/2)}{\sin(A/2)}$$1.5 = 2 \cos(A/2)$$\cos(A/2) = 1.5 / 2 = 0.75$Since $\sin(48^{\circ} 36^{\prime}) = 0.75$,we know that $\cos(90^{\circ} - 48^{\circ} 36^{\prime}) = 0.75$,which is $\cos(41^{\circ} 24^{\prime}) = 0.75$.Therefore,$A/2 = 41^{\circ} 24^{\prime} = 41.4^{\circ}$.Thus,$A = 2 	imes 41.4^{\circ} = 82.8^{\circ}$.

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 . . . . . . . $(\sin 48^{\circ} 36^{\prime} = 0.75)$

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