InterviewSolution
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InterviewSolution
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A cube has refractive index mu_(1) . There is a plane of refractive index mu_(2)(mu_(2) lt mu_(1)) upon it. A ray travelling through air is incident on a side face of the cube. The refracted ray is incident on the upper face of the cube at the minimum angle for total internal reflection to occur. Finaily the reflected ray emerges from opposite face. Show that if the angle of emergenceis phi then sin phi = sqrt(mu_(1)^(2) - mu_(2)^(2)). |
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Answer» Solution :In Fig. 2.28, PQ = incident ray on the side face of the cube, QR = refracted ray inside the cube, RS = totally reflected ray from the upper face of the cube, ST = emergent ray from the opposite face. Let critical ANGLE for total reflection be `theta_(c)`. According to the question, `theta._(c) approx theta_(c)` `i.e. "" sintheta._(c) approx theta_(c) = (1)/(2^(mu)1) = (1)/((mu1)/(mu_(2))) = (mu_(2))/(mu_(1))` Angle of incidence of the ray `RS= i = 90^(@) - theta_(c)` and angle of refraction= `PHI`. `THEREFORE "" (sini)/(sinphi) ` = refractive index of air with respect to the cube `= (1)/("refractive index of the with respect to air")` `or. "" (sini)/(sinphi) = (1)/(mu_(1))` `or, " " sinphi = mu_(1)sini` `= mu_(1) sin(90^(@) - theta_(c)) = mu_(1)costheta_(c)` `= mu_(1) sqrt(1 - sin^(2)theta_(c))` `= mu_(1) sqrt((1- mu_(2)^(2))/(mu_(1)^(2))) = sqrt(mu_(1)^(2) - mu_(2)^(2))` |
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