Problem 24.35: The lens of the eye. The crystalline lens of the human eye is a d

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Problem 24.35: The lens of the eye.

The crystalline lens of the human eye is a double-convex lens made of material having an index of refraction of 1.44 (although this varies). Its focal length in air is about 8.00 mm , which also varies. We shall assume that the radii of curvature of its two surfaces have the same magnitude.
(Note: The results obtained in the parts A, B and C are not strictly accurate, because the lens is embedded in fluids having refractive indexes different from that of air.)

Part A

Find the radii of curvature of this lens.

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Part B

If an object 11.0 cm tall is placed 32.0 cm from the eye lens, where would the lens focus it?

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Part C

How tall would the image be?

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Part D

Is this image real or virtual?

Please Choose

real

virtual

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Part E

Is it erect or inverted?

Please Choose

erect

inverted

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Problem 24.35: The lens of the eye.

The crystalline lens of the human eye is a double-convex lens made of material having an index of refraction of 1.44 (although this varies). Its focal length in air is about 8.00 mm , which also varies. We shall assume that the radii of curvature of its two surfaces have the same magnitude.
(Note: The results obtained in the parts A, B and C are not strictly accurate, because the lens is embedded in fluids having refractive indexes different from that of air.)

Part A

Find the radii of curvature of this lens.

R= mm

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Part B

If an object 11.0 cm tall is placed 32.0 cm from the eye lens, where would the lens focus it?

s= mm

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Part C

How tall would the image be?

|y|= mm

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Part D

Is this image real or virtual?

Please Choose

real

virtual

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Part E

Is it erect or inverted?

Please Choose

erect

inverted

SubmitMy Answers

Explanation / Answer

Here ,

part a) let the radius of curvature is R

1/f = (n - 1) *(2/R)

1/8 = (1.44 -1) * (2/R)

solving for R

R = 7.04 mm

the radius of curvature is 7.04 mm

part b)

let the distance is di

Using lens formula

1/f = 1/di + 1/do

1/.704 = 1/32 + 1/di

di = 0.719 cm = 7.19 mm

the lens would focus it at 7.19 mm

part c)

height of image = 11 * 0.709/(32)

height of image = 0.247 cm

height of image = 2.47 mm

part D)

as di is positive

the image is real

part E)

as di is positive

the image is inverted

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