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?

What would be the minimum power and associated focal length of the lens?

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?

Explanation / Answer

for minimum power
R1 = +10 mm

R2 = -6 mm

n = 1.43

Use Lens maker's formula,
1/f = (n-1)*(1/R1 - 1/R2)

= (1.43 - 1)*(1/10 + 1/6)

f_min = 8.72 mm

for maximum power
R1 = +6.5 mm

R2 = -5.5 mm

n = 1.43

Use Lens maker's formula,
1/f = (n-1)*(1/R1 - 1/R2)

= (1.43 - 1)*(1/6.5 + 1/5.5)

f_max = 6.93 mm

so,

minimum power = 1/f_min = 1/(8.72*10^-3) = 114.7 diaptor <<<<<<---------------Answer

maximum power = 1/f_max = 1/(6.93*10^-3) = 144.3 diaptor <<<<<<---------------Answer

given

object distnace, u = 27.5 cm

focal length for maximum power, f = 6.93 mm = 0.693 cm

Apply, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/0.693 - 1/27.5

v = 0.71 cm or 7.1 mm <<<<<<---------------Answer

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