An object is placed 12.5 cm in front of the cornea. (The cornea is thin ans has

Question

An object is placed 12.5 cm in front of the cornea. (The cornea is thin ans has approximately parallel sides so that the reflection that occurs as light travels from air to cornea to aqueous humor is essentially the same as though the aqueous humor were directly in contact with the air. The aqueous humor has index of refraction n = 1.34 and the radius of curvature of cornea is 7.8 mm.)

(a) What is the image distance for the image formed by the cornea alone?

I have this answer, it is 3.72 cm (including because I am sure this is used to find b)

(b) The image formed by the cornea serves as an object for the lens. Treat the lens as a thin lens 4 mm behind the cornea. Find the optical power of the lens necessary to form an image on the retina, 28 mm from the center of the lens.

Explanation / Answer

SOLUTION:

A)

If you use; n1/p + n2/q = (n2 - n1)/R

n1 = 1, p = 12.5 cm, n2 = 1.34, R = 0.078 cm

Solving for q, I got approximately 3.72 cm

B) The 28mm is your new image. And you know that the image from the cornea at 3.72 cm is now your new object, but it's not at 3.72 cm anymore because that's not the distance to the lens, it's the distance to the cornea. You have a 4mm offset to deal with. As for the sign, this image might help you. It might be best to draw it all out.

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