Question

SOLUTION SET UP The center of curvature of the first surface of the lens is on the outgoing side, so R = +6.0 mm. The center of curvature for the second surface is not on the outgoing side, so R2-_5.5 mn. We solve for f and then use the result in the thin-lens equation. Now we'll apply the thin-lens equation to the eye When light enters your eye, most of the focusing happens at the interface between the air and the cornea (the outermost element of the eye). The eye also has a double-convex lens, lying behind the conea, that completes the job of forming an image on the retina. (The lens also enables us to shift our distance of focus; it gets rounder for near vision and flatter for far vision.) The crystalline lens has an index of refraction of about 1.40. (a) For the lens shown (Figure 1), find the focal length. (b) If you could consider this lens in isolation from the rest of the eye, what would the image distance be for an object 0.20 m in front of this lens? SOLVE Part (a): Using the equation below, we find that (n-1)(4-斋) = R1 R2 Part (b): The object distance is do = 0.20 m-200 mm. Using the thin-lens equation Vf = 1/do + 1/ds,we obtain = 200min 72mm di = 7.5 mm REFLECT The image is slightly farther from the lens than it would be for an infinitely distant object. As expected for a converging lens, the focal length is positive ▼ Part A-Practice Problem: Suppose that to observe an object that is 5.69 m away, the lens in your eye flattens out so that it acts like the crystalline lens in the example, except with radii of curvature of G = 11.0 mm and C2-10.5 mm. How far behind the lens would the image form in this case? Express your answer to three significant figures and pay attention to units. Submit Request Answer Provide Feedback Next > gure 1 of 1 5.5 mm 6.0 mm C2 n= 1.40

Answer 1

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