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A light ray moving at a 54.3 deg
angle in water (n = 1.33) hits a
boundary with oil (n = 1.52). At
what angle is it refracted in the oil?
(Water n = 1.33, Air n = 1.00)
(Unit = deg)​

Respuesta :

opudodennis

Answer:

[tex]45.25^{\circ}[/tex]

Explanation:

Applying Snell's law

[tex]n_1sin\theta_1 = n2*sin\theta_2\\1.33*sin(54.3) = 1.52*sin\theta_2\\sin\theta_2 = 1.33*0.81208353/1.52 = 0.71057\\\theta_2=sin^{-1} (0.71057)\\\theta_2 = 45.25^{\circ}[/tex]

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