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Focus of the eye. 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?

Short Answer

Expert verified
(a) Radius of curvature needed is 0.875 cm. (b) It focuses text in front of retina. (c) Actual focus of mountain is in front, showing need for lens.

Step by step solution

01

Understanding the Problem

We need to determine the radius of curvature for the cornea that will focus parallel light rays from a distant object onto the retina, taking the index of refraction into account. For part (b), we will analyze if the same curvature can focus nearby objects like text on a screen. For part (c), we will consider if the actual physical curvature of the cornea properly focuses distant objects.
02

Using Lensmaker's Equation (Part a)

We're assuming the cornea behaves like a lens. The lensmaker's equation for a lens with one side in air is: \( \frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \). Here, for the cornea, we have \(R_2 = \infty\) since the back side is assumed flat, so \( \frac{1}{R_2} = 0 \). We need to find \(R_1 = R\) such that the focal length \(f = 25 \text{ mm} = 2.5 \text{ cm}\). We substitute these values in the equation: \[ \frac{1}{2.5} = (1.35 - 1) \left( \frac{1}{R} \right) \].
03

Solving for Radius of Curvature (Part a)

Simplifying the equation: \( \frac{1}{2.5} = 0.35 \times \frac{1}{R} \), therefore, \(R = 2.5 \times 0.35 = 0.875 \text{ cm} \). Thus, the radius of curvature must be \(0.875 \text{ cm}\) to focus a distant mountain on the retina.
04

Understanding Focal Capabilities (Part b)

We check if a screen 25 cm away would be focused by calculating the image distance using the thin lens equation: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), substituting \(f = 2.5 \text{ cm}\) and \(d_o = 25 \text{ cm}\). Thus \( \frac{1}{2.5} = \frac{1}{25} + \frac{1}{d_i} \).
05

Solving for Image Distance of Text (Part b)

Calculating from the equation: \( \frac{1}{d_i} = \frac{1}{2.5} - \frac{1}{25} \). Therefore, \( \frac{1}{d_i} = 0.4 - 0.04 = 0.36 \), resulting in \(d_i = \frac{1}{0.36} \approx 2.78 \text{ cm}\). The text focuses at 2.78 cm, in front of the retina.
06

Analyzing the Actual Curvature (Part c)

Using the real radius of curvature \(R = 0.5 \text{ cm}\), we calculate where it focuses the mountain. Plugging into the lens equation: \( \frac{1}{f} = 0.35 \times \frac{1}{0.5} = 0.7 \), thus \(f = \frac{1}{0.7} \approx 1.43 \text{ cm}\). So, it focuses in front of the retina at 1.43 cm, suggesting additional focusing power (lens help) is needed.

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Key Concepts

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

Lensmaker's Equation
The lensmaker's equation is a fundamental principle in optics that helps us understand how lenses form images. In the context of the human eye, it approximates the cornea as a lens, because the cornea bends light to help focus images on the retina.
The equation is presented as:
  • \( \frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \)
Here:
  • \( f \) is the focal length,
  • \( n \) is the refractive index of the lens material (for the eye, it's the index of the aqueous humor behind the cornea),
  • \( R_1 \) and \( R_2 \) are the radii of curvature of the two surfaces of the lens.
When considering the cornea:
  • The back surface is considered flat, making \( R_2 = \infty \), and thus \( \frac{1}{R_2} = 0 \).
By substituting these values, the equation simplifies, enabling us to solve for the radius of curvature \( R_1 \) that focuses images perfectly on the retina. Understanding and applying this equation helps us determine if a lens (or cornea) is appropriately shaped for clear vision.
Refraction
Refraction is the bending of light as it passes from one medium to another. This principle is critical in optics because it explains how lenses, including our eye’s cornea, can focus light.
In the human eye, when light enters through the cornea, it is bent or refracted so that it can be focused properly on the retina to form a clear image. The amount of bending is determined by the refractive index of the material.
The refractive index is a measure of how much light slows down in a medium, compared to in a vacuum. The refractive index of the aqueous humor in the eye is about 1.35, which bends light enough for the cornea to function effectively as a lens.
This bending is crucial in everyday vision:
  • It affects how images are focused, which determines the clarity of what we see.
  • Any problems in refraction can lead to common vision issues such as myopia or hyperopia.
Therefore, understanding refraction helps us appreciate how light behaves in our eyes and how corrective lenses work to address refractive errors.
Radius of Curvature
The radius of curvature is a measure of the curvature of a lens surface. For the cornea, this is a critical factor in determining how well it can focus light onto the retina.
In practical terms:
  • A smaller radius means a more curved surface, which bends light rays more sharply.
  • A larger radius indicates a flatter surface, bending light less.
For the eye's cornea, the ideal radius of curvature ensures that parallel rays from distant objects converge directly on the retina. This is crucial for clear vision of distant mountains, for example.
When the radius is not optimal:
  • Light may focus in front of or behind the retina, causing blurry vision.
In our exercise, determining the right radius of curvature using the lensmaker's equation is key to understanding how the eye achieves focus and why sometimes additional optical correction is necessary.

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Most popular questions from this chapter

A concave mirror is to form an image of the filament of a headlight lamp on a screen 8.00 m from the mirror. The filament is 6.00 mm tall, and the image is to be 36.0 \(\mathrm{cm}\) tall. (a) How far in front of the vertex of the mirror should the filament be placed? (b) To what radius of curvature should you grind the mirror?

If you place a concave mirror with a focal length of 1 \(\mathrm{m}\) into a liquid that has an index of refraction of \(3,\) what will happen? A. The mirror will no longer be able to focus light. B. The focal length of the mirror will decrease. C. The focal length of the mirror will increase. D. Nothing will happen.

\(\cdot\) The left end of a long glass rod 8.00 \(\mathrm{cm}\) in diameter and with an index of refraction of 1.60 is ground and polished to a convex hemispherical surface with a radius of 4.00 \(\mathrm{cm} .\) An object in the form of an arrow 1.50 \(\mathrm{mm}\) tall, at right angles to the axis of the rod, is located on the axis 24.0 \(\mathrm{cm}\) to the left of the vertex of the convex surface. Find the position and height of the image of the arrow formed by paraxial rays inci- dent on the convex surface. Is the image erect or inverted?

A spherical, concave shaving mirror has a radius of curva- ture of 32.0 \(\mathrm{cm}\) . (a) What is the magnification of a person's face when it is 12.0 \(\mathrm{cm}\) to the left of the vertex of the mirror? (b) Where is the image? Is the image real or virtual? (c) Draw a principal-ray diagram showing the formation of the image.

A double-convex thin lens has surfaces with equal radii of curvature of magnitude 2.50 \(\mathrm{cm} .\) Looking through this lens, you observe that it forms an image of a very distant tree, at a distance of 1.87 \(\mathrm{cm}\) from the lens. What is the index of refrac- tion of the lens?

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