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A ray of light is incident grazing the refracting surface of a prism having refracting angle A. It emerges the other refracting surface making an angle theta with the normal to the surface. Prove that the refractive index of the material of the prism is given by mu = [1+ ((cosA + sintheta)/(sinA))^(-2)]^(1//2). |
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Answer» Solution :For grazing incidence on the refracting surface AB of the PRISM, `i_(1) = 90^(@)` [Fig. 2.65]. So, in this case `r_(1) = theta_(c)` (critical angle) `"Now", " "A = r_(1) + r_(\2) or, r_(2) = A - r_(1) = A - theta_(c)` Considering REFRACTION at the second refracting surface, AC `"we get", mu = (sintheta)/(sinr_(2))` `or, " " sintheta = mu sinr_(2) = mu sin(A - theta_(c))` `= mu [sinA costheta_(c) - cosA sintheta_(c)]` `"But" " " sintheta_(c) = (1)/(mu) and costheta_(c) = sqrt(1- (1)/(mu^(2))) = (sqrt(mu^(2)-1))/(mu)` `therefore "" sintheta= mu [sinA* (sqrt(mu^(2) - 1))/(mu) -cos A * (1)/(mu)]` `= sin A sqrt(mu^(2) - 1)` `or, " " sintheta+cosA = sinAsqrt(mu^(2)-1)` `or, " " (sintheta + cosA)/(A) = sqrt(mu^(2) - 1)` `or, "" mu^(2) - 1 = ((sintheta + cos A)/(sinA))^(2)` `or, "" mu^(2) = 1 + ((sintheta + cos A )/(sinA))^(2)` `or, "" mu = [ 1 + ((sintheta + cos A )/(sinA ))^(2)]^(1/2)` |
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