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Solutions: Medical Option - the EYE

Q3.

A person has a myopic (short sighted) eye with a range of clear vision at distances from his eye of 0.15 m to 0.80m.

(i) Calculate the power of the correcting lens which would allow this eye to produce focused images of distant objects.

By 'distant objects' the examiner means objects at infinity.

The spectacle wearer would therefore need to focus objects at infinity at his/her far point. Therefore image distance would be 0.80 m. The image produced would be on the same side of the lens as the object and upright - so it would be a virtual image.

u = infinity so 1/u = zero

v = - 0.80 m

P = 1/f = 1/v + 1/u

P = - 1/0.80 + 0

P = - 1.25 D (must be negative!)

(ii) Calculate the new near point position for the eye when using the correcting lens.

When viewing a close up object the specatacle-wearer will need to form a virtual image of the object at his/her near point.

v = - 0.15

P = -1.25 D

P = 1/f = 1/u + 1/v

-1.25 = 1/u - 1/0.15

1/u = -1.25 + 1/0.15 = 5.42

u = 0.18 m

(Total 4 marks)