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(D) The refractive index $\mu$ of a prism is given by the formula:$\mu = \frac{\sin \left( \frac{A + \delta_m}{2} \right)}{\sin \left( \frac{A}{2} \ri

Vedclass · Accepted Answer

$1.414$

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(D) The refractive index $\mu$ of a prism is given by the formula:$\mu = \frac{\sin \left( \frac{A + \delta_m}{2} \right)}{\sin \left( \frac{A}{2} \right)}$Given:Refracting angle $A = 60^\circ$Angle of minimum deviation $\delta_m = 30^\circ$Substituting these values into the formula:$\mu = \frac{\sin \left( \frac{60^\circ + 30^\circ}{2} \right)}{\sin \left( \frac{60^\circ}{2} \right)}$$\mu = \frac{\sin(45^\circ)}{\sin(30^\circ)}$Since $\sin(45^\circ) = \frac{1}{\sqrt{2}}$ and $\sin(30^\circ) = \frac{1}{2}$:$\mu = \frac{1/\sqrt{2}}{1/2} = \frac{2}{\sqrt{2}} = \sqrt{2}$$\mu \approx 1.414$

$A$ prism of refracting angle $60^\circ$ is made with a material of refractive index $\mu$. For a certain wavelength of light,the angle of minimum deviation is $30^\circ$. For this wavelength,the value of the refractive index of the material is:

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