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Contact lenses are placed right on the eyeball, so the distance from the eye to an object (or image) is the same as the distance from the lens to that object (or image). A certain person can see distant objects well, but his near point is \(45.0 \mathrm{~cm}\) from his eyes instead of the usual \(25.0 \mathrm{~cm}\). (a) Is this person nearsighted or farsighted? (b) What type of lens (converging or diverging) is needed to correct his vision? (c) If the correcting lenses will be contact lenses, what focal length lens is needed and what is its power in diopters?

Short Answer

Expert verified
The person is farsighted and a converging lens is needed to correct his vision. The focal length of the lens required is 0.5625 m and the power of the lens required is approximately +1.75 diopters.

Step by step solution

01

Determine the type of eye defect

The person can see distant objects clearly but cannot see nearby objects clearly(i.e., objects closer than 45.0 cm, which is farther than the normal near point of 25.0 cm). This indicates that the person is farsighted. Farsightedness, also known as hypermetropia, is a condition where the incoming light is focused behind the retina, instead of directly on the surface.
02

Determine the type of lens required

To correct farsightedness, a converging lens is needed. The converging lens refracts the light rays inward such that they are focused directly onto the retina.
03

Calculate the Focal length and power of the lens

The object distance for the person is 45.0 cm and the normal near point distance is 25.0 cm. Using the lens formula, \[1/f = 1/v - 1/u\], where 'f' is the focal length, 'v' is the image distance and 'u' is the object distance, we get \[1/f = 1/25 - 1/45 = 4/225 cm^{-1}\] Now to convert it to meters, \[f = 1/(4/225) = 56.25 cm\] which is equal to \[0.5625 m\] We know that power (P) is the reciprocal of the focal length (in meters). Therefore, \[P = 1/f = 1/0.5625 = 1.78 diopters\] The power of the lens is usually rounded off to the nearest half-diopter, so the power of the contact lens needed is approximately +1.75 D.

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Key Concepts

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

Converging Lens
A converging lens, also known as a convex lens, curves outward on both sides and has the ability to bend (or converge) light rays towards a focal point. This characteristic allows converging lenses to correct farsightedness, a common vision defect where individuals struggle to see objects that are close to them.

When looking at nearby objects, the lens in a farsighted person’s eye doesn't bend light correctly, resulting in images being focused behind the retina. To rectify this, a converging lens is used in glasses or contact lenses. It adds the necessary convergence to light so the image falls directly on the retina for clear vision. Using a contact lens provides the benefit of placing the corrective power directly on the eye, ensuring that the distance from the eye to an object remains consistent, regardless of where the person is looking.
Lens Formula
The lens formula is a powerful tool in optics that relates the focal length of a lens to the distances of the object and the image from the lens. Mathematically, it is expressed as \[1/f = 1/v - 1/u\] where \(f\) is the focal length, \(v\) is the distance from the lens to the image, and \(u\) is the distance from the lens to the object.

For individuals with farsightedness, the lens formula helps calculate the required focal length of a lens that will bring the image from their far near point closer to the normal near point, thus allowing them to see nearby objects clearly. By knowing the near point of the eye with the vision defect, we can use the lens formula to find the focal length of the correctional lens.
Diopters in Optics
In the field of optics, diopters are units of measurement used to express the refractive power of a lens. One diopter represents the power of a lens with a one-meter focal length. The formula to calculate the power (P) in diopters is given by: \[P = 1/f\], where \(f\) is the focal length in meters.

For those who require vision correction, understanding the concept of diopters is crucial as it accurately describes the strength of lenses needed to correct their vision. Since the power of lenses is typically reported in increments of half diopters, the calculated diopter value is often rounded to the nearest 0.25 or 0.50 to match commercially available lens powers.
Vision Defects
Vision defects, or refractive errors, occur when the eye cannot focus light on the retina properly, leading to blurred vision. Farsightedness (hyperopia) is one such condition where distant objects are seen more clearly than nearby objects. This defect arises because the eyeball is too short or the cornea has too little curvature, causing light to focus behind the retina.

Other common refractive errors include nearsightedness (myopia), where close objects are clear and distant ones are blurred due to the opposite focal error; astigmatism, where irregularities in the cornea's curvature cause distorted or blurred vision at all distances; and presbyopia, the age-related difficulty in focusing on close objects. Each of these conditions requires specific corrective lenses that alter how light is bent into the eye, restoring clear vision.

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Most popular questions from this chapter

A tank whose bottom is a mirror is filled with water to a depth of \(20.0 \mathrm{~cm}\). A small fish floats motionless \(7.0 \mathrm{~cm}\) under the surface of the water. (a) What is the apparent depth of the fish when viewed at normal incidence? (b) What is the apparent depth of the image of the fish when viewed at normal incidence?

The cornea of the eye has a radius of curvature of approximately \(0.50 \mathrm{~cm},\) and the aqueous humor behind it has an index of refraction of \(1.35 .\) The thickness of the cornea itself is small enough that we shall neglect it. The depth of a typical human eye is around \(25 \mathrm{~mm}\). (a) What would have to be the radius of curvature of the cornea so that it alone would focus the image of a distant mountain on the retina, which is at the back of the eye opposite the cornea? (b) If the cornea focused the mountain correctly on the retina as described in part (a), would it also focus the text from a computer screen on the retina if that screen were \(25 \mathrm{~cm}\) in front of the eye? If not, where would it focus that text: in front of or behind the retina? (c) Given that the cornea has a radius of curvature of about \(5.0 \mathrm{~mm}\), where does it actually focus the mountain? Is this in front of or behind the retina? Does this help you see why the eye needs help from a lens to complete the task of focusing?

A candle \(4.85 \mathrm{~cm}\) tall is \(39.2 \mathrm{~cm}\) to the left of a plane mirror. Where is the image formed by the mirror, and what is the height of this image?

A frog can see an insect clearly at a distance of \(10 \mathrm{~cm}\). At that point the effective distance from the lens to the retina is \(8 \mathrm{~mm} .\) If the insect moves \(5 \mathrm{~cm}\) farther from the frog, by how much and in which direction does the lens of the frog's eye have to move to keep the insect in focus? (a) \(0.02 \mathrm{~cm}\), toward the retina; (b) \(0.02 \mathrm{~cm}\), away from the retina; (c) \(0.06 \mathrm{~cm}\), toward the retina; (d) \(0.06 \mathrm{~cm}\), away from the retina.

To determine the focal length \(f\) of a converging thin lens, you place a \(4.00-\mathrm{mm}\) -tall object a distance \(s\) to the left of the lens and measure the height \(h^{\prime}\) of the real image that is formed to the right of the lens. You repeat this process for several values of \(s\) that produce a real image. After graphing your results as \(1 / h^{\prime}\) versus \(s\), both in \(\mathrm{cm}\), you find that they lie close to a straight line that has slope \(0.208 \mathrm{~cm}^{-2}\). What is the focal length of the lens?

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