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In a simplified model of the human eye, the aqueous and vitreous humors and the lens all have a refractive index of \(1.40,\) and all the bending occurs at the cornea, whose vertex is \(2.60 \mathrm{~cm}\) from the retina. What should be the radius of curvature of the cornea such that the image of an object \(40.0 \mathrm{~cm}\) from the cornea's vertex is focused on the retina?

Short Answer

Expert verified
With the provided parameters for distance to the object and distance from the cornea to the retina, the equation is used first to give the focal length. Then this result is incorporated with the index of refraction into the lens maker's equation to find the radius of curvature for the cornea.

Step by step solution

01

Understand an Image Formation

To find the radius of curvature that will focus the light on the retina, you should first understand how an image is formed by a lens (or in this case, the cornea). Considering only refraction at the cornea, we can treat the cornea as a converging lens.
02

Use the Lens Equation

Next, we should use the Lens equation to solve the problem. The len's equation is: \(1/f = 1/d_0 + 1/d_i\) where \(f\) is the focal length of the lens, \(d_0\) is the object distance (distance from the object to the lens or cornea in this case), and \(d_i\) is the image distance (distance from the image to the lens). Given that all alternating happens at the cornea, we can consider the 'lens' to be at the front of the eye. Therefore, \(d_i = 2.6 cm\) (the distance from the cornea to the retina) and \(d_0= 40.0 cm\) (the distance to the object). Plugging these into the Lens equation gives a value for \(f\).
03

Convert Focal Length To Radius of Curvature

In order to find the radius of curvature, we must convert the focal length (\(f\)) to radius of curvature (\(R\)). The relationship between the focal length and radius of curvature is given by the Lensmaker's equation. For a single refractive surface, the equation is: \(1/f = (n-1)(1/R)\) where \(n\) is the index of refraction. Therefore, given that the index of refraction of the cornea is 1.40, plugging the earlier calculated \(f\) and given \(n\) into this equation will yield the value for the radius of curvature (\(R\)).

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Key Concepts

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

Refractive Index
In the field of optics, the refractive index is a measure of how much light bends, or refracts, when it enters a different medium. It is defined as the ratio of the speed of light in a vacuum to its speed in another medium. For example, the refractive index ( ") of the human eye components like the aqueous and vitreous humors, as well as the lens, is typically around 1.40. This means that light travels 1.40 times slower in these eye components than in a vacuum.

Refractive indices are crucial because they influence how images are formed and focused within optical systems, such as the eye. A higher refractive index indicates more significant bending of light rays. This property is vital in designing lenses to ensure that images converge properly on specific surfaces, like the retina.
Lensmaker's Equation
The Lensmaker’s Equation is an essential principle in optics used to relate the focal length of a lens to its physical characteristics. In its simplest form for a lens with uniform curvature, it's expressed as: \( \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \), where \(f\) is the focal length, \(n\) is the refractive index of the lens material, and \(R_1\) and \(R_2\) are the radii of curvature of the lens surfaces.

For single-surface refraction, as with a simple cornea, it's simplified to: \( \frac{1}{f} = (n - 1) \frac{1}{R} \). This equation helps in calculating the radius of curvature needed for an object to be focused correctly on the retina, ensuring a clear image.
Radius of Curvature
The radius of curvature is a measure of how curved a lens or surface is. Specifically, it's the radius of the spherical surface that forms the lens or part of the cornea responsible for focusing light. In optics, the curvature significantly affects how strongly the lens converges or diverges light rays.

For the human eye, understanding the radius of curvature of the cornea is vital. This curvature determines how the lens focuses light onto the retina. A precise radius of curvature ensures that images form on the retina, providing clear vision. It can be calculated through the Lensmaker's Equation, considering the refractive index and the focal length required for image formation.
Image Formation
Image formation in optics involves the convergence or divergence of light rays to create a visible replication of an object. This process is guided by the nature and curvature of lenses and the refractive index of materials involved.

In the context of the human eye, image formation occurs primarily at the cornea, where light first bends. The goal is for light rays from an object to converge accurately on the retina, forming a clear and sharp image. The cornea operates like a converging lens, redirecting light to form an image at a specific point, which is essential for effective vision.
Lens Equation
The lens equation is a fundamental formula in optics that relates the focal length of a lens to the object and image distances. It is usually written as: \( \frac{1}{f} = \frac{1}{d_0} + \frac{1}{d_i} \), where \(f\) is the lens's focal length, \(d_0\) is the distance from the object to the lens, and \(d_i\) is the distance from the lens to the image.

This equation is crucial for understanding how adjustments in the position of an object or the distance between the lens and the image affect where light will focus. By manipulating this equation, one can predict the focus and clarity of images in lenses used in various optical devices, including the human eye. This enables precise design and correction of lenses to achieve the desired image quality.

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Most popular questions from this chapter

A camera with a 90-mm-focal-length lens is focused on an object \(1.30 \mathrm{~m}\) from the lens. To refocus on an object \(6.50 \mathrm{~m}\) from the lens, by how much must the distance between the lens and the sensor be changed? To refocus on the more distant object, is the lens moved toward or away from the sensor?

A lens forms a real image that is \(214 \mathrm{~cm}\) away from the object and \(1 \frac{2}{3}\) times its height. What kind of lens is this, and what is its focal length?

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 thin lens with a focal length of \(6.00 \mathrm{~cm}\) is used as a simple magnifier. (a) What angular magnification is obtainable with the lens if the object is at the focal point? (b) When an object is examined through the lens, how close can it be brought to the lens? Assume that the image viewed by the eye is at the near point, \(25.0 \mathrm{~cm}\) from the eye, and that the lens is very close to the eye.

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?

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