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 114.7 diopters Number m What would be the minimum power and associated focal length of the lens? Number 144.3 diopters Number m At maximum power, how far behind the lens would the lens form an image of an object 27.5 cm in front of the front surface of the lens? Number 0.71 m

Explanation / Answer

apply the lens maker formula to find the focal length f   and

then use Power   = 1/f in meters

so lens maker equation is 1/f = (n-1) *(1/R1 -1/R2)

here n is refractive index = 1.430

R1 is radius ( for max power) = 5.5 mm

R2 is radius ( for max power) = 6.5 mm

so

1/f = ( 1.43-1) *(1/0.0055 -1/0.0065)

1/f = 12.02

f = 1/12.02 = 0.083 m    or 8.3 cm

so Power P   = 1/0.083   = + 12.02 Diopters

focal length f = 0.083 m

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for minimum power

R1 = 6 mm

R2 = 10 mm

so

1/f = (1.43-1) *(1/0.006 - 1/0.01)

1/f = 28.66

f = 0.034 m   or 3.4 cm

Power P = +28.67 Diopters
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use 1/f = 1/u + 1/v

for max power

f = 8.3 cm

u = 27.5 cm

so 1/v = 1/f - 1/u

v = uf/(u-f)

v = (27.5*8.3)/(27.5 -8.3)

v = 11.88 cm is the 0.11m -------<<<<<<<<<<<<Answer

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