My friend Lisa admitted that she needed reading glasses when her near point got

Question

My friend Lisa admitted that she needed reading glasses when her near point got to be 60 cm. What power lens, in diopters, would she need to be able to see clearly at a more comfortable distance of 25 cm? A very myopic person has a near point of 8 cm. and a far point of 20 cm. What power lens does he need to see clearly at a distance? What is his new near point with this vision correction? The rod and cone cells in the central part of the retima - the fovea - are packed more closely, giving a more detailed view. This area of increased rod and cone density has a diameter of about 1.5 mm. When you read a book, you want the image of the text you are reading to fall on the fovea. If you hold a book 50 cm from your eyes, how wide is the spot on the page whose image just fills the fovea? That is, how wide an area do you see most clearly? Assume that the image on the retina forms 17 mm from the lens in the eye (which gives 1/s' = 60.)

Explanation / Answer

Example 1:

Given;

Near point distance, v = - 60 cm;

Object distance, u = 25 cm;

From thin lens equation,

1/u + 1/v = 1/f

1/f = 1/25 cm + 1/-60 cm

f = 42.85 cm

The power of the lens is, P = 1/ f = 1/ 42.8 cm

P = 1/ 0.428 m

P = + 2.33 diopters

Example 2:

The original near point is at 8 cm.

The original far point is at 20 cm.

Part a)

As the person is myopic, to correct the short sightedness, a diverging lens should be used so that the virtual image should be formed at the original far point.

Thin lens equation:

1/f = 1/ - 1/(20 cm)

f = -20 cm

f = -0.2 m

The power of the lens, P = 1/f = 1/(-0.2)

P = +5 D

b) The new near point is the position for which the object will produce a virtual image at the original near point

1/(-20 cm) = 1/u - 1/(8 cm)

u = 13.33 cm

The new Near point with the correction in Part a) is 13.33 cm.

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