The cornea of the eye has a radius of curvature of approximately 0.40 cm , and the aqueous humor behind it has an index of refraction of 1.35. the thickness of the cornea itself is small enough that we shall neglect it. the depth of a typical human eye is around 25.0 mm. what would have to be the radius of curvature of the cornea so that it alone would focus the image of a distant mountain on the retina, which is at the back of the eye opposite the cornea?

0.64814 cm

Explanation:

= Refractive index of air = 1

= Refractive index of aqueous humor = 1.35

u = Object distance =

v = Image distance = 25 mm

R = Radius of curvature of the cornea

Lens equation

Radius of curvature of the cornea is 0.64814 cm

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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0.11 m/s and 0.28 m/s should be correct