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Derive the formula for Refractive index of prism in case of minimum deviation condition

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Refractive index of prism (\mu )in case of minimum deviation condition-

As we learned The angle of deviation $(\delta)$ for the prism is given as  \ \delta =i+e-A

and from The plot of \delta \ \ vs \ \ i we get i=e \ \ then \ \ \delta =\delta _{min}

 i.e \ \ \delta_{min} =i+e-A=i+i-A=2i-A\\\Rightarrow i=\frac{A+\delta _{min}}{2}

For the prism of refractive index \mu places in the air.

For the first surface, we can write 1*sini=\mu sinr_1

similarly For the second surface, we can write \mu sinr_2=1* sine

using i=e we get r_1=r_2

 A=r_1+r_2=2r_1\\ \rightarrow r_1=\frac{A}{2}

So 1*sini=\mu sinr_1 will give us

\Rightarrow 1*sin(\frac{A+\delta _{min}}{2})=\mu sin(\frac{A}{2})\\ \Rightarrow \mu =\frac{sin(\frac{A+\delta _{min}}{2})}{ sin(\frac{A}{2})}

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avinash.dongre

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