A typical human lens has an index of refraction of 1.430. The lens has a double

Question

A typical human lens has an index of refraction of 1.430. The lens has a double convex shape, but its curvature can be varied by the ciliary muscles acting around its rim. At minimum power, the radius of the front of the lens is 10.0 mm while that of the back is 6.00 mm. At maximum power the radii are 6.50 mm and 5.50 mm, respectively. If the lens were in air, what would be the maximum power and associated focal length of the lens? Number Number diopters Im What would be the minimum power and associated focal length of the lens? Number Number diopters Im At maximum power, how far behind the lens would the lens form an image of an object 15.5 cm in front of the front surface of the lens? Number Im

Explanation / Answer

Given double convex lens in air with

index of refraction n = 1.430

and radius of curvature R1 = 10 mm , R2 = 6 mm

we know that the power of the lens is P = 1/f

and focal length is 1/f = (n-1) (1/R1 -1/R2)

substituting the values  

1/f1 = (1.43-1)(1/10 -1/6)

1/f1 = -0.02867*10^-2 diopters = P1

f1 = -34.88 mm

and

1/f2 = (1.43-1)(1/6.5 -1/5.5)

1/f1 = -0.012028*10^-2 = P2

f2 = -83.14 mm

for maximum power the focal length is f1 = -34.88 mm

object distance is s = 15.5 cm ,image distance is  

1/s' = 1/f - 1/s

1/s' = -1/34.88 -1/155

s' = -28.47 mm

the image in front of the surface of the lens is at 28.47 mm

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