The indices of refraction for her contact lens, cornea, and the fluid behind her cornea are 1.6, 1.4, and 1.3, respectively. Light is incident from air onto her contact lens at an angle of 30 ∘∘ from the normal of the surface. At what angle is the light traveling in the fluid behind her cornea?
Answer:
[tex]23^{\circ}[/tex]
Explanation:
n = Refractive index of air = 1
[tex]n_1[/tex] = Refractive index of contact lens = 1.6
[tex]n_2[/tex] = Refractive index of cornea = 1.4
[tex]n_3[/tex] = Refractive index of fluid = 1.3
From Snell’s law
[tex]n\sin30^{\circ}=n_1\sin\theta\\\Rightarrow \theta=\sin^{-1}\dfrac{1\sin30^{\circ}}{1.6}\\\Rightarrow \theta=18.21^{\circ}[/tex]
[tex]n_1\sin\theta=n_2\sin\theta_1\\\Rightarrow \theta_{1}=\sin^{-1}\dfrac{1.6\times \sin18.21^{\circ}}{1.4}\\\Rightarrow \theta_1=20.92^{\circ}[/tex]
[tex]n_2\sin\theta_1=n_3\sin\theta_3\\\Rightarrow \theta_3=\sin^{-1}\dfrac{1.4\sin20.92^{\circ}}{1.3}\\\Rightarrow \theta_3=22.62^{\circ}\approx 23^{\circ}[/tex]
The angle is the light traveling in the fluid behind her cornea is [tex]23^{\circ}[/tex].