A wave is traveling through air at an angle of 75° before it enters an eye lens cover. If air has an index of refraction of 1.00, and an eye lens cover has an index of refraction of 1.38, at what angle is the wave traveling once it enters the eye lens cover?

Respuesta :

Answer:

[tex]44.4^o[/tex]

Explanation:

Use Snell`s law,

[tex]n_i\ sin\theta_i=n_r\sin\theta_r[/tex]

Here [tex]\theta_i=75^o[/tex] is the angle of incidence, [tex]\theta_r=?[/tex] angle of refraction,[tex]n_i=1, n_r=1.38[/tex] are the refractive indexes of first and second medium.

Substitute the given values, we get

[tex]1\times sin75^o=1.38\times sin\theta_r[/tex]

[tex]0.6999=sin\theta_r[/tex]

[tex]\theta_r=sin^-^1(0.6999)=44.4^o[/tex]

Thus, at [tex]44.4^o[/tex] angle the wave traveling once it enters the eye lens cover.