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Calculate critical angle of incidence between two material with different refractive indices n1 = 1.4 and n2 = 1.36. Also calculate numerical aperture and acceptance cone angle.
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a) Critical angle $(\phi _ c) = Sin^{-1}(\frac {n_2}{n_1}) \\ \phi_c = Sin^{-1}(1.36/1.4) \\ \phi_c = 76.27^o $

b) Numerical Aperture(NA) $= (n1^2-n2^2)^{1/2} \\ NA = (1.4^2-1.36^2)^{1/2} \\ NA = 0.33$

C) Acceptance angle $ (\phi_a) = Sin^{-1} NA \\ \phi _a = Sin^{-1}0.33 \\ \phi_a = 19.32^O$

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