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Derive expression for refractive index of prism

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A=Prism angle, δ=Angle of deviation, i1​=Angle of incidence, i2​=Angle of emergent.

 

In the case of minimum deviation,∠r1​=∠r2​=∠r

 

A=∠r1​+∠r2​

SO, A=∠r+∠r=∠2r

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\angle r=\frac{A}{2}$ \\Now, again $A+\delta=i_{1}+i_{2}(\because$ In the case of minimum deviation $i_{1}=i_{2}=i$ and $\delta=\delta_{m}$ \\$So , A+\delta_{m}=i+i=2 i$ \\$Now , i=\frac{A+\delta_{m}}{2}$ \\Now, from snell's rule, $\mu=\frac{\sin i}{\sin r}

\mu=\frac{\sin \frac{A+\delta_{m}}{2}}{\sin \frac{A}{2}}

 

Posted by

Deependra Verma

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