What is the angle of refraction in a medium if the angle of incidence in air (n=1.00) is 48˚ and the index of refraction of the medium is 1.58?
sin48°sin x°=n2n1→x=sin−1(sin48°n2)=sin−1(sin48°1.58)=28°\frac{\sin48°}{\sin\ x°}=\frac{n_2}{n_1}\to x=\sin^{-1}(\frac{\sin48°}{n_2})=\sin^{-1}(\frac{\sin48°}{1.58})= 28°sin x°sin48°=n1n2→x=sin−1(n2sin48°)=sin−1(1.58sin48°)=28°
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