Question 5 of 8 Map E UNIVERSITY PHYSICS A typical human lens has an index of re

Question

Question 5 of 8 Map E UNIVERSITY PHYSICS 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 ciliarymuscles 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? diopters What would be the minimum power and associated focal length of the lens? Number diopters At maximum power, how far behind the lens would the lens form an image of an object 29.5 cm in front of the front surface of the lens? Number A Previous 2 Check Answer O Next H Exit O Hint

Explanation / Answer

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

= 6.93*10^-3 m <<<<<-------------Answer

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

= 8.72*10^-3 m<<<<<-------------Answer

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

given

object distnace, u = 21.5 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 - 1/29.5

v = 0.71 cm

= 7.1*10^-3 m <<<<<<---------------Answer

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