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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 cm from the retina. What should be the radius of curvature of the cornea such that the image of an object 40.0 cm from the cornea's vertex is focused on the retina?

Short Answer

Expert verified
The radius of curvature of the cornea should be approximately 0.71 cm.

Step by step solution

01

Understand the Problem Statement

We need to find the radius of curvature of the cornea required to focus an image on the retina. The cornea is acting as a lens with refractive index 1.40, and the object distance from the cornea is 40.0 cm while the image distance (to the retina) is 2.60 cm.
02

Use Lens Maker's Equation for a Refractive Surface

The lens maker's equation for a single refractive surface is given by:\[\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}\]where \(n_1 = 1.00\) (refractive index of air), \(n_2 = 1.40\) (refractive index of the cornea), \(u = -40.0\) cm (object distance negative since it's a real object), and \(v = 2.60\) cm (image distance). We need to solve this for \(R\).
03

Substitute Known Values Into the Equation

Plug the values into the lens maker's equation:\[\frac{1.40}{2.60} - \frac{1.00}{-40.0} = \frac{1.40 - 1.00}{R}\]
04

Simplify the Equation

Calculate the left side of the equation:\[\frac{1.40}{2.60} + \frac{1.00}{40.0}\]Calculate these fractions individually and then add them to find:\[0.5385 + 0.025 = 0.5635\]
05

Solve for the Radius of Curvature \(R\)

Now solve \(0.5635 = \frac{0.40}{R}\) for \(R\):\[R = \frac{0.40}{0.5635} \approx 0.71 \text{ cm}\]
06

Finalize the Answer

Thus, the radius of curvature of the cornea should be approximately 0.71 cm for the object 40 cm away to be focused on the retina.

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Key Concepts

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

Lens Maker's Equation
The lens maker's equation is a key tool in optics that describes how lenses form images. It relates the focal length of a lens to its curvature and refractive index. In a simplified form for a single refractive surface, the equation is written as: \( \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \) where:- \( n_1 \) is the refractive index of the medium from which light is coming, usually air, which is 1.00.- \( n_2 \) is the refractive index of the lens material, such as the cornea, which in this problem is 1.40.- \( u \) is the object distance from the lens, typically taken as negative for real objects as they are on the opposite side of the incoming light.- \( v \) represents the image distance, or where the image forms.- \( R \) is the radius of curvature of the lens's surface. This equation helps in calculating how curved a lens needs to be to focus an object at a specific distance onto a point, like the retina in the eye.
Refractive Index
Refractive index is a measure of how much light slows down as it passes through a medium. It's represented by the symbol \( n \) and is calculated as the ratio of the speed of light in a vacuum to the speed of light in the medium. For any material:- When \( n = 1 \), light travels through without slowing down. This is true for a vacuum or air at standard conditions.- When \( n > 1 \), light slows down. The greater the refractive index, the slower the light travels through the medium. This is typical for water, glass, and biological tissues.For the human eye, both the cornea and the lens have a refractive index of around \( 1.40 \). This value is vital as it determines how much the eye can bend light, ultimately focusing images clearly on the retina. Understanding this property helps optometrists design lenses that correct vision.
Radius of Curvature
The radius of curvature refers to the radius of the sphere from which a lens's surface is derived. It's a critical factor that affects how lenses converge or diverge light signals. In simple terms, it's the distance from the lens's surface center to the center of that sphere. For lenses: - A large radius means the lens surface is less curved and bends light less sharply. - A small radius indicates a strongly curved surface that drastically changes the light’s path. In our human eye model, adjusting the cornea's radius of curvature is crucial. It should be just right to ensure that the image of an object lands perfectly on the retina. An accurate computation ensures clear vision, as seen in the problem where the corneal curvature required is approximately 0.71 cm. This precision aligns perfectly for distant objects like those 40 cm away.
Human Eye Model
The human eye model simplifies how light enters and is focused within our eyes. In basic terms, the eye is like a camera, where many elements work closely to form sharp images. The main components of this model include: - **Cornea**: It's the eye's front surface that starts the process of focusing light. With a refractive index of 1.40, it provides most of the eye's optical power. - **Lens**: Further refines the focus, adjusting as needed to clearly see near and far objects. - **Aqueous and Vitreous Humors**: These are gel-like substances in the eye that help maintain its shape and guide light to the lens and retina. By using the lens maker's equation in our simplified model of the human eye, we can calculate how the cornea should bend light for it to hit the retina correctly. This model offers essential insights for understanding how we see and how vision can be corrected with glasses.

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Most popular questions from this chapter

The eyepiece of a refracting telescope (see Fig. 34.53) has a focal length of 9.00 cm. The distance between objective and eyepiece is 1.20 m, and the final image is at infinity. What is the angular magnification of the telescope?

A concave mirror has a radius of curvature of 34.0 cm. (a) What is its focal length? (b) If the mirror is immersed in water (refractive index 1.33), what is its focal length?

The focal length of the eyepiece of a certain microscope is 18.0 mm. The focal length of the objective is 8.00 mm. The distance between objective and eyepiece is 19.7 cm. The final image formed by the eyepiece is at infinity. Treat all lenses as thin. (a) What is the distance from the objective to the object being viewed? (b) What is the magnitude of the linear magnification produced by the objective? (c) What is the overall angular magnification of the microscope?

A person swimming 0.80 m below the surface of the water in a swimming pool looks at the diving board that is directly overhead and sees the image of the board that is formed by refraction at the surface of the water. This image is a height of 5.20 m above the swimmer. What is the actual height of the diving board above the surface of the water?

The cornea behaves as a thin lens of focal length approximately 1.8 cm, although this varies a bit. The material of which it is made has an index of refraction of 1.38, and its front surface is convex, with a radius of curvature of 5.0 mm. (a) If this focal length is in air, what is the radius of curvature of the back side of the cornea? (b) The closest distance at which a typical person can focus on an object (called the near point) is about 25 cm, although this varies considerably with age. Where would the cornea focus the image of an 8.0-mm-tall object at the near point? (c) What is the height of the image in part (b)? Is this image real or virtual? Is it erect or inverted? (\(Note:\) The results obtained here are not strictly accurate because, on one side, the cornea has a fluid with a refractive index different from that of air.)

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