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The uncorrected far point of Colin's eye is \(2.0 \mathrm{m} .\) What refractive power contact lens enables him to clearly distinguish objects at large distances?

Short Answer

Expert verified
Colin needs contact lenses with a refractive power of \(-0.5\) diopters.

Step by step solution

01

Understand the Far Point

The far point is the maximum distance at which the eye can see objects clearly without accommodation. In this problem, Colin's far point is given as 2 meters.
02

Understand Key Formula

To find the refractive power of the contact lens, we can use the lens formula: \( \frac{1}{f} = P \), where \(f\) is the focal length and \(P\) is the refractive power in diopters. For lenses correcting myopia (nearsightedness), the far point is treated as the focal length.
03

Use the Lens Formula

We have \( f = -2.0 \) m (negative because the focal length for myopia is negative). Insert into the formula: \( P = \frac{1}{-2.0} \). Calculate \( P \).
04

Perform Calculation

Calculate the refractive power: \( P = \frac{1}{-2.0} = -0.5 \) diopters. The negative sign indicates a diverging lens is needed.
05

Conclusion

Colin’s contact lens should have a refractive power of \(-0.5\) diopters to correct his vision for seeing distant objects clearly.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Far Point
The concept of the far point is crucial in understanding how our vision works. It's the maximum distance at which the eye can see objects clearly without needing to adjust its focal power. For people with normal eyesight, the far point is considered to be at infinity, meaning they can see objects far away without effort.
When myopia, or nearsightedness, is present, the far point is much closer. In Colin's case, his far point is measured at 2 meters. This means that without corrective lenses, he cannot see objects clearly beyond this distance.
Understanding the far point allows us to determine the necessary corrective measures to bring distant objects into focus, which is vital in optical physics.
Refractive Power
Refractive power refers to a lens's ability to bend light, bringing rays into focus. It's expressed in diopters, a unit that measures the focusing ability of a lens. The refractive power is directly related to how effective a lens will be in correcting vision.
The formula used to find refractive power is \[ P = \frac{1}{f} \]where \( f \) is the focal length of the lens in meters. Positive diopters are used for lenses that converge light, such as those correcting farsightedness. Meanwhile, negative diopters, like those needed for Colin's myopia, indicate diverging lenses that spread out light rays to extend the far point.
Refractive power is a fundamental concept in optical physics that helps tailor vision correction to individual needs.
Myopia Correction
Myopia, commonly known as nearsightedness, occurs when the eye's shape causes light rays to focus in front of the retina when viewing distant objects. This makes far objects appear blurry.
To correct myopia, lenses with negative refractive power are used. These lenses are known as diverging or concave lenses. They help spread the light rays outward, effectively extending the eye's far point back to infinity.
The goal is to adjust the focus so that light rays fall precisely on the retina when looking into the distance. By using a corrective lens with the correct refractive power, such as the \(-0.5\) diopter lens required for Colin, the individual's vision is restored for distant viewing.
Lens Formula
The lens formula is essential for determining the needed specifications to correct vision problems. It provides a straightforward way to calculate the refractive power required for a lens.\[ \frac{1}{f} = P \]This basic yet powerful formula connects the focal length \( f \) of a lens with its refractive power \( P \). For a person with myopia, the focal length corresponding to their uncorrected far point is used.In Colin's situation, his far point was 2 meters, meaning the focal length needed is \(-2.0\) m, leading to the calculation:\[ P = \frac{1}{-2.0} = -0.5 \text{ diopters} \]
This tells us a diverging lens of \(-0.5\) diopters will adjust the focus of objects back to infinity, then allowing Colin to view distant objects with clarity.
Mastering the lens formula is invaluable in optical physics and ensures precision in vision correction.

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Most popular questions from this chapter

Use the thin-lens equation to show that the transverse magnification due to the objective of a microscope is \(m_{o}=-U f_{o} .\) [Hints: The object is near the focal point of the objective: do not assume it is at the focal point. Eliminate \(p_{0}\) to find the magnification in terms of \(q_{0}\) and $\left.f_{0} . \text { How is } L \text { related to } q_{0} \text { and } f_{0} ?\right].$
A camera with a 50.0 -mm lens can focus on objects located from $1.5 \mathrm{m}$ to an infinite distance away by adjusting the distance between the lens and the film. When the focus is changed from that for a distant mountain range to that for a flower bed at \(1.5 \mathrm{m},\) how far does the lens move with respect to the film?

A converging lens with a focal length of \(15.0 \mathrm{cm}\) and a diverging lens are placed \(25.0 \mathrm{cm}\) apart, with the converging lens on the left. A 2.00 -cm-high object is placed \(22.0 \mathrm{cm}\) to the left of the converging lens. The final image is \(34.0 \mathrm{cm}\) to the left of the converging lens. (a) What is the focal length of the diverging lens? (b) What is the height of the final image? (c) Is the final image upright or inverted?

A nearsighted man cannot clearly see objects more than \(2.0 \mathrm{m}\) away. The distance from the lens of the eye to the retina is \(2.0 \mathrm{cm},\) and the eye's power of accommodation is \(4.0 \mathrm{D}\) (the focal length of the cornea-lens system increases by a maximum of \(4.0 \mathrm{D}\) over its relaxed focal length when accommodating for nearby objects). (a) As an amateur optometrist, what corrective eyeglass lenses would you prescribe to enable him to clearly see distant objects? Assume the corrective lenses are $2.0 \mathrm{cm}$ from the eyes. (b) Find the nearest object he can see clearly with and without his glasses.
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