Problem 39 - College Physics

Question

\(\cdot\) The lens of the eye. The crystalline lens of the human eye is a double-convex lens made of material having an index of refraction of 1.44 (although this varies). Its focal length in air is about 8.0 \(\mathrm{mm}\) , which also varies. We shall assume that the radii of curvature of its two surfaces have the same magnitude. (a) Find the radii of curvature of this lens. (b) If an object 16 \(\mathrm{cm}\) tall is placed 30.0 \(\mathrm{cm}\) from the eye lens, where would the lens focus it and how tall would the image be? Is thisimage real or virtual? Is it erect or inverted? (Note: The results.obtained here are not strictly accurate, because the lens is embedded in fluids having refractive indexes different from that of air.)

Step-by-Step Solution

Answer

(a) Radius of curvature \(= 0.704 \text{ cm}\). (b) Image is at \(0.816 \text{ cm}\), real, inverted, \(-0.435 \text{ cm}\) tall.

Step 1: Understanding Lens Formula

In optics, the relationship between the object distance \(d_o\), image distance \(d_i\), and focal length \(f\) of a lens is given by the lens formula: \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\). We'll use it later to find the image distance.

Step 2: Calculate Radius of Curvature

The lens maker's equation, which relates the focal length \(f\), the refractive index \(n\), and the radii of curvature \(R_1\) and \(R_2\) of a lens, is \(\frac{1}{f} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\). For a symmetric lens, \(R_1 = -R_2 = R\). Therefore, \(\frac{1}{f} = (n - 1)\frac{2}{R}\). Solving for \(R\), we have \(R = 2(n - 1)f\). Substituting \(n = 1.44\) and \(f = 0.8\, \text{cm}\) yields \(R = 2(1.44 - 1)(0.8) = 0.704\, \text{cm}\).

Step 3: Calculate Image Position

Using the lens formula \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\) with \(f = 0.8\, \text{cm}\) and \(d_o = 30.0\, \text{cm}\), we can calculate \(d_i\). Rearrange to \(\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o}\). Substituting gives \(\frac{1}{d_i} = \frac{1}{0.8} - \frac{1}{30}\). Solving gives \(d_i = 0.816\, \text{cm}\).

Step 4: Image Characteristics

Using magnification \(m\) formula \(m = -\frac{d_i}{d_o}\), calculate \(m = -\frac{0.816}{30}\). The image is \(m imes 16\, \text{cm} = -0.435\, \text{cm}\) tall, meaning it's inverted and smaller. The negative sign indicates the image is real and inverted, as it forms on the opposite side of the lens from the object.

Key Concepts

Crystalline Lens

The crystalline lens is a vital component of the human eye. This lens is a unique, clear structure that helps focus light onto the retina, allowing us to see images clearly. It is a flexible, doubly convex structure, meaning it bulges outward from both sides. The flexibility of the crystalline lens is key to its ability to change shape, a process known as accommodation. This adjustment enables the eye to focus on objects at various distances, achieving clarity whether the object is near or far.

The lens helps refine and direct light to the fovea, the part of the retina that provides the sharpest vision. Over time, the lens can become less flexible and less clear, which can affect one's vision, often leading to conditions like presbyopia or cataracts.

Overall, the crystalline lens is crucial for achieving clear, focused vision and plays a pivotal role in the complex process of sight.

Refractive Index

The refractive index is a measure that indicates how much a substance, such as the crystalline lens, can bend light. Each material has its own refractive index, which depends on how fast light travels through it compared to the speed of light in a vacuum. In the context of the human eye, it's important because it directly affects how light is focused onto the retina.

For the crystalline lens, the refractive index varies slightly but is often around 1.44. This means that light travels 1.44 times slower in the lens material than it does in air. A higher refractive index indicates that the lens is more effective at bending and focusing light.

Light refracting through the lens allows the eye to focus on objects at various depths, from distant landscapes to nearby reading material. Understanding refractive index is essential for fields like optics and ophthalmology, as it influences the design of corrective lenses and the analysis of visual systems.

Focal Length

The focal length of a lens is the distance from the lens to the point where incoming parallel light rays converge to a single point or appear to diverge from. In the human eye, this point is crucial in determining where images are focused on the retina.

For the crystalline lens, the focal length in air is approximately 8.0 mm. However, this can vary depending on several factors, such as the curvature and elasticity of the lens and the refractive index of the surrounding medium. The focal length is adjustable through accommodation, allowing the eye to change focus between near and far objects.

This functionality is essential for ensuring objects at various distances can be viewed sharply and with clarity. In optical physics, understanding focal length helps in calculating image formation and is instrumental in developing eyeglasses, contact lenses, and other vision-improving devices.

Lens Maker's Equation

The lens maker's equation is a fundamental formula in optics that relates the focal length of a lens to the refractive index and the radii of curvature of its surfaces. It is typically 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, and \(R_1\) and \(R_2\) are the radii of curvature for the two lens surfaces.

In the exercise, assuming a symmetric lens with equal and opposite radii of curvature simplifies this to \( \frac{1}{f} = (n - 1)\frac{2}{R} \), highlighting how these parameters collectively determine the focusing power of the lens. Solving for the radius of curvature involves using the known refractive index and focal length to indicate the specific curvature needed for desired lens performance.

Understanding this equation is imperative for lens crafting and design, as it informs the geometric and material specifications needed for lenses to perform effectively. It is the basis for designing anything from camera lenses to eyeglass lenses, ensuring they meet required optical standards.