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Chapter 33: Problem 95

For a person whose near point is \(115 \mathrm{~cm},\) so that he can read a computer monitor at \(55 \mathrm{~cm},\) what power of read ing glasses should his optician prescribe, keeping the lenseye distance of \(2.0 \mathrm{~cm}\) for his spectacles?

Short Answer

Expert verified
Answer: The optician should prescribe reading glasses with a power of approximately 1.31 diopters.

Step by step solution

01

Understand the lens formula

The lens formula for thin lenses is given by: $$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$ where f is the focal length, v is the image distance (near point), and u is the object distance (distance to the computer monitor).
02

Convert the near point and monitor distance to the same units

In the problem, we are given the near point in cm. We need to convert it to meters since most standard optics problems use meters. The near point is 115 cm, which is equal to 1.15 meters. The monitor distance is given as 55 cm, which is equal to 0.55 meters.
03

Find the image distance with the lense-eye distance

We are given the lense-eye distance as 2.0 cm, which is equal to 0.02 meters. To find the image distance (near point) considering the lense-eye distance, we can subtract the lense-eye distance from the desired monitor distance: $$v = 0.55 - 0.02 = 0.53 \text{ meters}$$
04

Use the lens formula to find the focal length

We can now plug the image distance (v) and object distance (u) into the lens formula to find the focal length (f): $$\frac{1}{f} = \frac{1}{0.53} - \frac{1}{1.15}$$ Solve for f: $$f = \frac{1}{(\frac{1}{0.53} - \frac{1}{1.15})} \approx 0.766 \text{ meters}$$
05

Calculate the lens power

The power (P) of a lens in diopters is the reciprocal of the focal length in meters: $$P = \frac{1}{f}$$ Plug the focal length into the formula: $$P = \frac{1}{0.766} \approx 1.305 \text{ diopters}$$ The optician should prescribe reading glasses with a power of approximately 1.31 diopters for the person to be able to read a computer monitor at a distance of 55 cm.

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Most popular questions from this chapter

Several small drops of paint (less than \(1 \mathrm{~mm}\) in diameter) splatter on a painter's eyeglasses, which are approximately \(2 \mathrm{~cm}\) in front of the painter's eyes. Do the dots appear in what the painter sees? How do the dots affect what the painter sees?

Will the magnification produced by a simple magnifier increase, decrease, or stay the same when the object and the lens are both moved from air into water?

Jack has a near point of \(32 \mathrm{~cm}\) and uses a magnifier of 25 diopter. a) What is the magnification if the final image is at infinity? b) What is the magnification if the final image is at the near point?

A person with a near-point distance of \(24.0 \mathrm{~cm}\) finds that a magnifying glass gives an angular magnification that is 1.25 times larger when the image of the magnifier is at the near point than when the image is at infinity. What is the focal length of the magnifying glass?

An object is placed on the left of a converging lens at a distance that is less than the focal length of the lens. The image produced will be a) real and inverted. c) virtual and inverted. b) virtual and erect. d) real and erect.
See all solutions

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