Problem 33 - University Physics with Modern Physics

Question

The Cornea As a Simple Lens. The cornea behaves as a thin lens of focal length approximately \(1.8 \mathrm{cm},\) 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 \(\mathrm{mm} .\) (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 \(25 \mathrm{cm},\) 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.)

Step-by-Step Solution

Answer

(a) \( R_2 \approx 1.86 \text{ cm} \). (b) \( d_i \approx 1.937 \text{ cm} \). (c) \( h_i \approx -0.062 \text{ cm} \), real and inverted.

Step 1: Calculate Equivalent Focal Length Using Lensmaker's Equation

The Lensmaker's equation for a lens surrounded by a medium, e.g. air, is given by:\[\frac{1}{f} = (n - 1)\left( \frac{1}{R_1} - \frac{1}{R_2} \right)\]where \( n = 1.38 \) is the refractive index of the cornea material, \( R_1 = 5.0\, \text{mm} = 0.5\, \text{cm} \) is the radius of curvature of the front surface, and \( R_2 \) is the unknown radius of curvature of the back surface.\ We need to find \( R_2 \) such that \( f = 1.8\, \text{cm} \). Rewriting:\[\frac{1}{1.8} = (1.38 - 1)\left( \frac{1}{0.5} - \frac{1}{R_2} \right)\]

Step 2: Solve for the Radius of Curvature of the Back Side

Solve the Lensmaker's equation for \( R_2 \):\[\frac{1}{1.8} = 0.38\left( \frac{1}{0.5} - \frac{1}{R_2} \right)\]First, calculate \( \frac{1}{0.5} = 2 \), then:\[\frac{1}{1.8} = 0.38\left( 2 - \frac{1}{R_2} \right)\]Multiply both sides by 1.8:\[1 = 0.684\left( 2 - \frac{1}{R_2} \right)\]Divide by 0.684:\[\frac{1}{0.684} = 2 - \frac{1}{R_2}\]Solve for \( \frac{1}{R_2} \) by isolating it:\[\frac{1}{R_2} = 2 - 1.462 \]\[\frac{1}{R_2} = 0.538\]Thus, \( R_2 = \frac{1}{0.538} \approx 1.86\, \text{cm} \). So, the back side radius is approximately \( 1.86\, \text{cm} \).

Step 3: Determine Image Distance for Near Point Object

Using the lens formula \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( d_o = 25\, \text{cm} \) is the object distance and \( f = 1.8\, \text{cm} \), solve for \( d_i \):\[\frac{1}{1.8} = \frac{1}{25} + \frac{1}{d_i}\]First, calculate \( \frac{1}{25} = 0.04 \), then:\[\frac{1}{1.8} = 0.04 + \frac{1}{d_i}\]Subtract 0.04 from both sides:\[\frac{1}{1.8} - 0.04 = \frac{1}{d_i}\]Convert \( \frac{1}{1.8} \approx 0.556 - 0.04 = 0.516 \), thus:\[\frac{1}{d_i} = 0.516\]So, \( d_i = \frac{1}{0.516} \approx 1.937\, \text{cm} \).

Step 4: Calculate Image Height and Determine Image Nature

Using the magnification formula \( m = \frac{h_i}{h_o} = -\frac{d_i}{d_o} \), where \( h_o = 0.8\, \text{cm} \) is the object height, find \( h_i \):\[h_i = h_o \left( -\frac{d_i}{d_o} \right)\]Substitute the values:\[h_i = 0.8 \times \left( -\frac{1.937}{25} \right) \approx -0.062\, \text{cm}\]The negative sign indicates the image is inverted. Since \( d_i \) is positive, the image is real.

Key Concepts

Focal Length

The focal length of a lens is the distance between the lens and its focus. This is where parallel rays of light converge after passing through the lens. In the case of the cornea, it acts as a lens with a given focal length of approximately 1.8 cm. This means that light after passing through the cornea will focus at a point that is 1.8 cm away from it.
The focal length is crucial as it dictates how a lens converges or diverges light. In lenses, the shorter the focal length, the more powerful the lens is, meaning it can bend light significantly.
In this exercise, understanding that the focal length is given allows us to use the Lensmaker's equation to find other properties like the radius of curvature of the lens' back surface.

Index of Refraction

The index of refraction, also known as the refractive index, is a measure of how much a substance slows down light compared to vacuum. It is defined by the speed of light in vacuum divided by the speed of light in the medium. For the cornea, this index is given as 1.38,
meaning that light travels 1.38 times slower in the cornea than in a vacuum.

• Refractive index tells us how much the path of light is bent, or refracted, as it enters the material.
• A higher index means more bending, which importantly contributes to the focal properties of the lens.

This property is used in the Lensmaker's equation, where you calculate the focal length based on refractive index and radii of curvature of the lens surfaces. Understanding the index helps us predict how light will behave when passing through different substances, such as from air to cornea.

Radius of Curvature

Radius of curvature refers to the radius of the sphere whose surface makes up the lens. For the cornea, the radius of curvature of its front surface is given as 5.0 mm, and we are tasked to find the back surface radius.
The curvature affects how much light the lens can bend:

• A smaller radius of curvature results in a more curved surface, which bends light rays more sharply.
• Conversely, a larger radius means a flatter surface, bending light less.

In this exercise, using Lensmaker’s equation helps find the back curvature needed to achieve the specified focal length. The exact value found using the formula provides insight into the geometry needed for the cornea to function correctly as part of the eye's focusing mechanism.

Object and Image Distance

The object distance (\(d_o\)) is the distance from the object to the lens, while the image distance (\(d_i\)) is from the lens to the formed image. For a functioning lens, these distances are fundamental in determining where an image will appear relative to the lens.
In solving lens problems, we use the lens formula:
\[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]Given that the object is at 25 cm, corresponding to the near point for typical vision, this position helps find where the image forms.

• If \(d_i\) is positive, the image is real and on the opposite side of where light enters the lens.
• If negative, the image is virtual, appearing on the same side as the object.

In the solution provided, \(d_i\) is positive, hence the image is real, indicating the lens indeed focuses the light to form a tangible image on the retina, for instance.