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Derive the expression for refractive index of the material of the prism in terms of angle of the prism and angle of minimum deviation. |
Answer» Solution : From `Delta MQR, ( i- r_(1)) + ( e - r_(R)) = delta` So `( I + e ) - ( r_(1) + r_(2)) = delta ` From`Delta PQN , r_(1) + r_(2) + /_ QNR = 180^(@)` Also `/_ A + /_N = 180^(@)` Thus `/_ A = r_(1) + r_(2)` So, `i + e - A = delta ` At minimum deviation, `i = e, r_(1) = r_(2) = r `and `delta = dleta _(m)` `RARR i = ( A + delta_(m))/( 2)` And `r = ( A )/( 2)` Also `mu = ( sin i )/( sin r )` Hence`mu = ( sin ((A + delta_(m))/( 2)))/( sin ((A)/(2)))` |
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