The cornea of the eye has a radius of curvature of approximately 0.500 cm, and t

Question

The cornea of the eye has a radius of curvature of approximately 0.500 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 2.41 cm.

A)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? (Note: The eye is acting like a spherical surface not a thin lens.) R = ? cm

B)If the cornea focused the mountain correctly on the retina as described in part A, where would it focus the text from a computer screen if that screen were 26.6 in front of the eye? i = ? cm

C)Given that the cornea has a radius of curvature of about 0.450 , where does it actually focus the mountain? i = ? cm

Explanation / Answer

The cornea of the eye has a radius of curvature  is  R_1  = 0.500cm aqueous humor behind it has an index of refraction of  n  = 1.35 The depth of a typical human eye is around R_2 = 2.41 cm.  From lens maker formula   n_1 / d_0 + n_2 / d_i  = n_2 - n_1 / R  1 /   +  1.35 / 2.41 cm  = ( 1.35 - 1) / R    Radius of curvature  R  = 0.624 cm    b)  From lens makers formula      n_1 / d_0 + n_2 / d_i  = n_2 - n_1 / R     1/ 26.6 cm + 1.35 / d_i  = 1.35 - 1 / 0.624 cm     image distance is  d_i  = 2.58 cm  image is focussed behind retina     c ) radius of curvature is   R  = 0.450cm    
         n_1 / d_0 + n_2 / d_i  = n_2 - n_1 / R           1/ 26.6 cm + 1.35 / d_i  = 1.35 - 1 / 0.450cm         d_i  = 1.82 cm  image forms in front of retina     1/ 26.6 cm + 1.35 / d_i  = 1.35 - 1 / 0.450cm         d_i  = 1.82 cm  image forms in front of retina       

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