Problem 32 - University Physics with Modern Physics

Question

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 this image 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) Radii of curvature are 0.704 cm. (b) Image is 0.82 cm from the lens, 0.4368 cm tall, real and inverted.

Step 1: Calculate Radii of Curvature Using Lensmaker's Equation

The Lensmaker's equation for a lens in air is \( \frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \). Given \( f = 8 \text{ mm} = 0.8 \text{ cm} \) and \( n = 1.44 \), assume that \( R_1 = R_2 = R \). Therefore, the equation simplifies to \( \frac{1}{0.8} = (1.44 - 1) \left( \frac{1}{R} + \frac{1}{R} \right) \). This becomes \( \frac{1}{0.8} = 0.44 \times \frac{2}{R} \). Solving for \( R \), we have \( R = 0.44 \times 2 \times 0.8 = 0.704 \text{ cm} \).

Step 2: Use the Lens Formula to Find Image Position

The lens formula is \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( d_o = 30 \text{ cm} \) is the object distance. With the given focal length \( f = 0.8 \text{ cm} \), substitute to get \( \frac{1}{0.8} = \frac{1}{30} + \frac{1}{d_i} \). Solving for \( d_i \), we find \( d_i = \frac{0.8 \times 30}{30 - 0.8} = \frac{24}{29.2} \approx 0.82 \text{ cm} \). The image is formed at \( 0.82 \text{ cm} \) on the opposite side of the lens.

Step 3: Determine Image Characteristics

The image distance is positive, which indicates the image is real and located on the opposite side of the lens from the object. According to physics of lenses, a real image formed by a single convex lens is inverted.

Step 4: Calculate Image Height Using Magnification

Magnification \( m \) is given by \( m = -\frac{d_i}{d_o} \). Plug in the known values \( d_i = 0.82 \text{ cm} \), \( d_o = 30 \text{ cm} \), so \( m = -\frac{0.82}{30} = -0.0273 \). The image height \( h_i \) is \( h_i = m \times h_o = -0.0273 \times 16 = -0.4368 \text{ cm} \). The negative sign indicates the image is inverted.

Key Concepts

Lensmaker's Equation

The Lensmaker’s equation is a fundamental tool in optics used to calculate the focal length of a lens in air. It posits that the focal length (\( f \)) of a lens is influenced by the radii of curvature of its two surfaces (\( R_1 \) and \( R_2 \)), and the refractive index (\( n \)) of the material from which the lens is made. The equation is usually given as:

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

In this case, the exercise assumes both surfaces have equal radii (\( R_1 = R_2 = R \)), simplifying the equation to \( \frac{1}{f} = 2(n - 1)/R \). By rearranging to solve for \( R \), we find the radius of curvature for the human eye's lens. Understanding this equation is key to grasping how changes in the lens’s shape or refractive index affect its focusing ability.

Focal Length

Focal length, a crucial concept in optics, is the distance from the center of a lens to the point where parallel rays of light converge or appear to diverge from. In simple terms, it indicates how strongly a lens can bend light. A shorter focal length means stronger light-bending power and a closer focal point.

• In the context of the human eye, the given focal length is 8 mm.
• This value represents how the eye focuses light to form images on the retina.

Calculating the focal length using practical measurements, like those in the exercise, helps us appreciate the precision of the eye's optical system. The ability to focus accurately is essential for clear vision, underlining the importance of balancing the focal length with the lens's geometry and the surrounding refractive environment, like fluids in the eye.

Image Formation

Image formation by lenses is described by the lens formula:

• \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \)

where \( d_o \) is the object distance, \( d_i \) is the image distance, and \( f \) is the focal length. This formula helps us determine the position where an image is formed relative to the lens.
The exercise shows how light from an object placed some distance away converges to form an image either within or beyond the focal length range. Analyzing this formation reveals whether the resulting image is real or virtual, and upright or inverted.

• A real image is produced on the opposite side of the lens, as is the case here, indicating that it can be projected onto a screen.
• In this scenario, the image is inverted, a characteristic of images formed by convex lenses when the object is beyond the focal length.

Understanding these fundamentals enhances one’s ability to predict how lenses modify light paths.

Refractive Index

The refractive index (\( n \)) of a medium is a measure of how much it reduces the speed of light compared to a vacuum. It is a crucial factor in optics because it determines how much light is bent, or refracted, when entering or exiting a material.

• In the exercise, the lens material has a refractive index of 1.44.
• This value signifies the optical density of the lens, affecting focal length and image quality.

A higher refractive index means that light slows down more as it passes through, refracting strongly. This bending ability is exploited in lenses to control light direction and image focus. Variations in the refractive index, due to changes in material or environmental conditions, can significantly alter the performance of optical systems like a human eye lens. Adjusting these factors allows for precise image correction and focus, essential in optical devices.