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Vinvika [58]
2 years ago
8

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?
Physics
1 answer:
Paladinen [302]2 years ago
4 0

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

<h3>What is Snell's law?</h3>

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.

brainly.com/question/10112549

#SPJ1

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