1. A far sighted person has a near point NP = 72.0 cm and a far point FP = 600 m

Question

1. A far sighted person has a near point NP = 72.0 cm and a far point FP = 600 m. State clearly what the object distance and image distance should be when considering correcting for this persons vision. What is the power (Diopters) of a correction lens for this person assuming that the person needs to hold a physics textbook 25 cm from her/his face for comfortable reading? Calculate the power assuming a contact lens correction. Is the correction lens converging or diverging? Show your steps completely for credit.

Explanation / Answer

here object distance s = 25 cm

image distance s' = -72 cm

1/f = 1/s + 1/s'

1/f = (1/25)-(1/72)

f = 38.29 cm

power p = 1/f ( in m)

P = 1/(0.3829)

P = + 2.612 D

P = +ve the lens in converging

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