absolute refractive index of water is 4/3 a ray of light passing from water to air form angle of incidence of 35 .find angle of refraction

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Solution

Using relation,
$n = \frac{\sin i}{\sin r} \frac{4}{3} = \frac{\sin 35^{0}}{\sin r} or \sin r = \frac{3}{4} \sin 35^{0} = \frac{3}{4} \times 0.57 = 0.43 r = \sin^{- 1} (0.43) = 49^{0}$

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absolute refractive index of water is 4/3 a ray of light passing from water to air form angle of incidence of 35 .find angle of refraction

Using relation, n=sin isin r43=sin 350sin ror sin r=34sin 350=34×0.57=0.43r=sin-1(0.43)=490

Using relation,
$n = \frac{\sin i}{\sin r} \frac{4}{3} = \frac{\sin 35^{0}}{\sin r} or \sin r = \frac{3}{4} \sin 35^{0} = \frac{3}{4} \times 0.57 = 0.43 r = \sin^{- 1} (0.43) = 49^{0}$