InterviewSolution
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InterviewSolution
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A beam of light of wavelength `400 nm` is incident normally on a right angled prism as shown in Fig. It is observed that light just grazes along the surface `AC` after falling on it. If refractive index `mu` of the material of prism varies with wavelength `lamda` as `mu = 1.2 + (b)/(lamda^2)` Calculate the value of `b and mu` of prism material for `lamda = 500 nm`. Given `theta = sin^-1 (0.625)`. . |
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Answer» As the ray goes grazingly along `AC, theta` must be the critical angle. From `mu = (1)/(sin C) = (1)/(sin theta) = (1)/(0.625) = 1.6` As `mu = 1.2 + (b)/(lamda^2)` `1.6 = 1.2 + (b)/((400)^2` `b = 0.4 xx (400)^2 = 64000 nm^2` For `lamda = 500 nm`, the refractive index of the material of the prism is `mu = 1.2 + (b)/(lamda^2) = 1.2 + (64000)/((500)^2) = 1.2 + 0.256` `mu = 1.456`. |
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