6 -5 points SerCP10 25.P009 My Notes Ask Your The near point of a person\'s eye

Question

6 -5 points SerCP10 25.P009 My Notes Ask Your The near point of a person's eye is 72.2 cm. (Neglect the distance from the lens to the eye.) (a) To see objects clearly at a distance of 24.0 cm, what should be the focal length of the appropriate corrective lens? (b) To see objects clearly at a distance of 24.0 cm, what should be the power of the appropriate corrective lens? diopters Need Help? ase -15 points SerCP10 25P My Notes O Ask Your An artificial lens is implanted in a person's eye replace a diseased lens. The distance between the artificial lens and the retina is 2.76 cm. In the absence of the lens, an image of a distant object (formed by refraction at the cornea) falls 5.09 cm behind the implanted lens. The lens is designed to put the image of the distant object on the retina what is the power of the implanted lens? Hint: Consider the image formed by the cornea to be a virtual object. diopters Need Help? SP My Notes Ask Y Your The desired overall magnification of a compound microscope is 141x. The objective alone produces a lateral magnification of 10.0x. Determine the required focal ength of the eyeplece Need Help?

Explanation / Answer

Answer 6)

Using the formula
1/f = 1/v + 1/u
f is the focal length, v is the image distance, u is the object distance
u = 24 cm ; v = -72.2 cm (negative because the image is virtual)

After substituting in the above formula we get

1/f = -1/72.2 + 1/24

Solving for f we get f = 0.359 mts

Power in dioptres is = 1/f = 2.78 D

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