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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?

Step by step solution

01

Understand the Lensmaker's Equation

For part (a), we need to understand the lensmaker's equation. Given that the index of refraction, \(n\), of the aqueous humor is 1.35, and the radius of curvature, \(R\), is \(0.50 \mathrm{cm}\), the equation that relates these is \[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]where \(R_1\) is the radius of curvature of the front surface and \(R_2\) the back surface. Assuming the back surface is flat, \(R_2\) becomes infinity and does not affect the equation.

02

Calculate the Focal Length for Part (a)

Substitute into the lensmaker's formula to find the focal length \(f\):\[ \frac{1}{f} = (1.35 - 1) \left( \frac{1}{0.5} \right) \]This simplifies to \( \frac{1}{f} = 0.7 \). Solving for \(f\), we get:\[ f = \frac{1}{0.7} \approx 1.4286 \mathrm{cm} \]

03

Calculate Desired Radius of Curvature for Perfect Focus (Part a)

To focus a distant object on the retina at the eye's depth \(25 \mathrm{mm} = 2.5 \mathrm{cm}\), we need\[ f = 2.5 \mathrm{cm} \]Use the lensmaker's formula again and solve for \( R \):\[ (1.35 - 1) \left( \frac{1}{R} \right) = \frac{1}{f} \]\[ 0.35 \times \frac{1}{R} = \frac{1}{2.5} \]\[ R = \frac{2.5}{0.35} \approx 7.14 \mathrm{cm} \]

04

Analyze Part (b) - Computer Screen Focusing

Using the focal length found in Step 2 (\( f = 1.4286 \mathrm{cm} \)), apply the lens formula \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]for an object distance \(d_o = 25 \mathrm{cm}\) (computer screen): \[ \frac{1}{1.4286} = \frac{1}{25} + \frac{1}{d_i} \]\[ 0.7 = 0.04 + \frac{1}{d_i} \]\[ \frac{1}{d_i} = 0.66 \Rightarrow d_i \approx 1.515 \mathrm{cm} \]Since \(d_i = 1.515 \mathrm{cm}\), which is less than the depth (2.5 cm), the image focuses in front of the retina.

05

Calculate Actual Image Distance for Part (c)

Given \(R = 5.0 \mathrm{mm} = 0.5 \mathrm{cm}\), use the lensmaker's formula to find where it focuses the mountain.\[ 0.35 \times \frac{1}{0.5} = \frac{1}{f} \Rightarrow 0.7 = \frac{1}{f} \Rightarrow f = 1.4286 \mathrm{cm} \]This focal length confirms that distant objects focus before the retina in a 2.5 cm deep eye, suggesting it's behind. Additional help from other parts of the eye or corrective lenses 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 powerful tool in optics that helps us determine how a lens will focus light. This equation considers various physical properties of the lens and its surrounding environment to calculate the focal length. The formula is:

• \( \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \)

Here:

• \( f \) is the focal length of the lens.
• \( n \) is the index of refraction of the material between lens surfaces.
• \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces.

When one radius of curvature, say \( R_2 \), represents a flat surface, it becomes infinity \((\infty)\), simplifying the equation. Understanding this equation is crucial when manipulating lenses in devices like eyeglasses or cameras.

Radius of Curvature

The radius of curvature refers to the curvature radius of a lens's surface. It essentially determines how curved the lens is. A smaller radius of curvature indicates a more sharply curved lens, whereas a larger radius corresponds to a flatter lens surface. In optical instruments, this property is vital because:

• It affects the focus and magnification power of the lens.
• Helps in determining the lens's ability to converge or diverge light.

In the human eye, the cornea's radius of curvature plays a crucial role in focusing light onto the retina. Its precise curvature helps determine whether the eye can naturally focus on distant objects or whether corrective lenses are needed.

Focal Length

The focal length of a lens is the distance over which parallel rays of light are brought to a focus by the lens. It is a fundamental measurement in optics because it describes how strongly the lens converges or diverges light.

• Shorter focal lengths mean a greater bending of the light, leading to a wider field of view.
• Longer focal lengths result in a narrower field of view and are thus used for zooming in on distant targets.

In the context of the human eye, the focal length must correspond to the distance from the cornea to the retina for clear vision. Any discrepancy necessitates optical correction.

Index of Refraction

The index of refraction, or refractive index, is a measure of how much a ray of light bends when it passes through a material. It's calculated using the ratio of the speed of light in a vacuum to that in the material. This property is vital for determining how lenses bend light rays.

• Materials with a higher refractive index bend light more than those with a lower refractive index.
• The refractive index is also crucial for the design of lenses in optics, affecting calculations like those in the lensmaker's equation.

For the human eye, the refractive index of the cornea plays a major role in focusing images accurately on the retina. When the natural refractive index fails to focus light correctly, vision correction through lenses or surgery is needed.