Problem 27.46 Part A A person\'s prescription for his new bifocal eyeglasses cal

Question

Problem 27.46 Part A A person's prescription for his new bifocal eyeglasses calls for a refractive power of -0.0600 diopters in the distance-vision part and a power of 1.20 diopters in the close-vision part. Assuming the glasses rest 2.00 cm from his eyes and that the corrected near-point distance is 25.0 cIm, determine the near and far points of this person's uncorrected vision. dear point dnear point IIn Submit My Answers Give Up Incorrect; Try Again; 7 attempts remaining Provide Feedback Continue

Explanation / Answer

Here, for the nearsighted correction f = 1/diopter = 1/(-0.0600) = -16.7 m

Again using the expression, 1/f = 1/di + 1/do we can find di (far point)

now do = infinity so 1/f = 1/di so di = f = -16.7 m

so the far point is 1670 cm + 2cm = 1672 cm = 16.72 m

For the farsighted f = 1/1.20 = 0.833 m = 83.3 cm with the lens we use 83.3 + 2 = 85.3 cm

Here we have, do = 25cm

So 1/di = 1/f - 1/do = 1/85.3 - 1/25 = -0.0283 => di = -35.4cm or 35.4 cm in front of the eye

Therefore, we have d(far point) = 16.72 m

and d(near point) = 35.4 cm.

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