Question

Most of the bending of light in the eye is done at the air-cornea interface. By what angle (θcornea) is the beam of light shown in Figure deviated as it passes from air to the cornea if the incident angle is θi = 23.6°? The refractive index of air is nair = 1.00, the refractive index of the cornea is ncornea = 1.38. Explain why θcornea< θi?

Answer 1

The angle (θcornea)  when light passes from air to cornea is 16.86°

What is Snell's law?

It states that the ratio of sine of angle of incidence and angle of refraction is equal to the refractive index of second medium to the  first medium.

sini/sinr =n₂ / n₁

Most of the bending of light in the eye is done at the air-cornea interface. The beam of light deviated as it passes from air to the cornea if the incident angle is θi = 23.6°.

Given the angle of incidence i = 23.6°, refractive index of air n₁ =1, refractive index of cornea n₂ = 1.38, then the angle of refraction at cornea is

sinr = sini x (n₁/n₂)

Plug the values, we get

sinr = sin23.6 x (1/1.38)

sinr = 16.86°

The angle of refraction is less than angle of incidence due to refraction.

Thus, the angle (θcornea)  when light passes from air to cornea is 16.86°

Learn more about Snell's law.

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