A prism is made of glass having a refractive | vedclass

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(B) Let $\delta_m$ be the angle of minimum deviation and $A$ be the angle of the prism.We know the formula for the refractive index $\mu$ of a prism i

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$90$

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(B) Let $\delta_m$ be the angle of minimum deviation and $A$ be the angle of the prism.We know the formula for the refractive index $\mu$ of a prism is given by:$\mu = \frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}$Given that $\mu = \sqrt{2}$ and $\delta_m = A$,we substitute these values into the formula:$\sqrt{2} = \frac{\sin \left(\frac{A+A}{2}\right)}{\sin \left(\frac{A}{2}\right)}$$\sqrt{2} = \frac{\sin A}{\sin \left(\frac{A}{2}\right)}$Using the trigonometric identity $\sin A = 2 \sin \left(\frac{A}{2}\right) \cos \left(\frac{A}{2}\right)$,we get:$\sqrt{2} = \frac{2 \sin \left(\frac{A}{2}\right) \cos \left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right)}$$\sqrt{2} = 2 \cos \left(\frac{A}{2}\right)$$\cos \left(\frac{A}{2}\right) = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$Since $\cos(45^{\circ}) = \frac{1}{\sqrt{2}}$,we have:$\frac{A}{2} = 45^{\circ}$$A = 90^{\circ}$

$A$ prism is made of glass having a refractive index of $\sqrt{2}$. If the angle of minimum deviation is equal to the angle of the prism,then the angle of the prism is: (in $^{\circ}$)

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