1. A far sighted person has a near point NP = 72.0 cm and a far point FP = 600 m
ID: 1292542 • Letter: 1
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.
2. A lens of focal length f = 6.00 mm is to be used as a magnifier. Where must an object be placed in order for the image to have maximum magnification? Is the image real or virtual? What is the magnification? Is the magnification positive or negative for a magnifier?
Explanation / Answer
object distance s = 25 cm
image distance s' = -72 cm (NP)
1/f = 1/s + 1/s'
1/f = (1/25)-(1/72)
f = 38.29 cm
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power P = 1/f ( in m) = 1/0.3829 = +2.61 D
converging lens
A)Obj placed @ focal length
B) Virtual img (when @ or inside focal length)
C)M= 25.0cm/f = 25/0.6 = 41.66 (repeating)
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