The refractive index of a transparent liquid | vedclass

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(A) The formula for the refractive index $\mu$ of a prism is given by $\mu = \frac{\sin((D_{\min} + A)/2)}{\sin(A/2)}$.Given that the prism is equilat

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$30$

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(A) The formula for the refractive index $\mu$ of a prism is given by $\mu = \frac{\sin((D_{\min} + A)/2)}{\sin(A/2)}$.Given that the prism is equilateral,the angle of the prism $A = 60^{\circ}$.The refractive index $\mu = \sqrt{2}$.Substituting these values into the formula:$\sqrt{2} = \frac{\sin((D_{\min} + 60^{\circ})/2)}{\sin(60^{\circ}/2)}$$\sqrt{2} = \frac{\sin((D_{\min} + 60^{\circ})/2)}{\sin(30^{\circ})}$Since $\sin(30^{\circ}) = 1/2$,we have:$\sqrt{2} = \frac{\sin((D_{\min} + 60^{\circ})/2)}{1/2}$$\frac{\sqrt{2}}{2} = \sin((D_{\min} + 60^{\circ})/2)$$\frac{1}{\sqrt{2}} = \sin((D_{\min} + 60^{\circ})/2)$Since $\sin(45^{\circ}) = 1/\sqrt{2}$,we get:$(D_{\min} + 60^{\circ})/2 = 45^{\circ}$$D_{\min} + 60^{\circ} = 90^{\circ}$$D_{\min} = 30^{\circ}$.

The refractive index of a transparent liquid filled in an equilateral hollow prism is $\sqrt{2}$. The angle of minimum deviation for the liquid will be (in $^{\circ}$)

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