Question 5 of 10 Incorrect Map A E UNIVERSITY PHYSICS Presented by sapling learn

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Question 5 of 10 Incorrect Map A E UNIVERSITY PHYSICS Presented by sapling learning 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 335 diopters 0.003 What would be the minimum power and associated focal length of the lens Number Number 0.004 267 diopters At maximum power, how far behind the lens would the lens form an image of an object 21.5 cm in front of the front surface of the lens? Number 2.94 Previous 3 Give Up & View Solution Check Answer Next Exit Hint

Explanation / Answer

Lens maker's formula

1/F = (ng/nm - 1)*(1/R1 - 1/R2)

ng = refractive indes of glass(eye) = 1.43

nm = refractive indes of medium = 1

R1 = radius of front surface

R1 = radius of back surface

F = focal length

for maximum power

R1 = 6.5 mm

R2 = -5.5 mm

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

F = 6.93 mm = 0.00693 m <<<-answer

pmax = 1/F(in m) = 1/( 0.00693) = 144.3 D <<<-------answer

++++++++++++

for minimum power

R1 = 10 mm

R2 = -6 mm

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

F = 8.72 mm = 0.00872 m <<<------answer

pmin = 1/F(in m) = 1/(0.00872) = 114.7 D <<-----------answer

+++++++++++++++++++++

from lens equation

1/s + 1/s' = 1/F

F = 0.693 cm

s = 25 cm

s' = ?

1/25 + 1/s' = 1/0.693

image distance s' = 0.713 cm

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