Problem 44 - University Physics with Modern Physics

Question

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 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 cm tall is placed 30.0 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 7.04 mm. (b) Image is focused 0.822 cm from lens, height is -0.438 cm, real and inverted.

Step 1: Use Lensmaker's Equation

To find the radii of curvature, we will use the Lensmaker's equation for a double-convex lens, which is given by \( \frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \), where \( f \) is the focal length, \( n \) is the index of refraction, and \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces. As given, \( f = 8.0 \) mm and \( n = 1.44 \). Also, since the radii have the same magnitude, \( R_1 = -R_2 = R \). Substituting these into the equation gives us: \[ \frac{1}{8} = (1.44-1) \left( \frac{1}{R} + \frac{1}{R} \right) \].

Step 2: Solve for Radius of Curvature

Substituting the known values into the equation, we have: \( \frac{1}{8} = 0.44 \times \frac{2}{R} \). Rearrange and solve for \( R \): \[ \frac{1}{8} = \frac{0.88}{R} \]. Therefore, \[ R = \frac{0.88 \times 8}{1} \]. Simplifying, we find \[ R = 7.04 \text{ mm} \].

Step 3: Use Lens Equation for Image Location

To find where the image will be focused, use the thin lens formula \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( d_o \) is the object distance and \( d_i \) is the image distance. Here, \( d_o = 30.0 \text{ cm} \) and \( f = 0.8 \text{ cm} \) (converted from mm to cm). Rearranging gives us \( \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} = \frac{1}{0.8} - \frac{1}{30} \).

Step 4: Calculate Image Distance

Substitute in the known values to find the image distance: \( \frac{1}{d_i} = 1.25 - 0.0333 \), which simplifies to \( \frac{1}{d_i} \approx 1.2167 \). Therefore, \( d_i \approx \frac{1}{1.2167} \approx 0.822 \text{ cm} \). This positive value indicates the image is real and formed on the opposite side of the lens from the object.

Step 5: Use Magnification to Find Image Height

The magnification \( m \) is given by \( m = -\frac{d_i}{d_o} \). Using the values \( d_i = 0.822 \text{ cm} \) and \( d_o = 30.0 \text{ cm} \), the magnification is \( m = -\frac{0.822}{30} \approx -0.0274 \). The image height \( h_i \) is then \( h_i = m \times h_o \) with \( h_o = 16 \text{ cm} \). Thus, \( h_i = -0.0274 \times 16 \approx -0.438 \text{ cm} \). The negative sign indicates the image is inverted.

Key Concepts

Crystalline Lens

The crystalline lens, found in the human eye, is a flexible, transparent structure responsible for focusing light and forming images on the retina. Much like a camera lens, it adjusts the focal length to focus on objects at various distances, enhancing our ability to see clearly. Its unique double-convex shape and elasticity allow it to change shape, a process known as accommodation. This adaptability is crucial for tasks like reading or looking at distant objects.

The structure of the crystalline lens comprises multiple layers, including a softer outer capsule and firmer inner fibers. This combination contributes to its refractive abilities, helping to direct light precisely onto the retina.

In optical physics, understanding the crystalline lens aids in solving problems related to image formation, as it behaves like a standard convex lens, utilizing principles such as the lensmaker's equation.

Index of Refraction

The index of refraction, often denoted as \( n \), is a measure of how much light slows down as it passes through a material. For the crystalline lens, this index typically ranges from 1.38 to 1.42, depending on hydration and protein content, though in our exercise example, we used 1.44.

This value is crucial because it affects how the lens bends light. A higher index indicates that light will slow down more, bending it further as it enters a denser medium. This bending is what helps focus light rays to form clear images.

Materials with a high index of refraction are often used in optical applications to create strong focusing effects. Understanding how different indices affect light travel is key for applying concepts like the lensmaker's equation to predict lens behavior.

Radii of Curvature

The radii of curvature (\( R_1 \) and \( R_2 \)) of a lens describe the curvature of its surfaces. When we say a lens is double-convex, it means both the front and back surfaces are convex, or outwardly curved. In the provided example, both surfaces have equal curvature, which simplifies calculations.

The radii affect the lens's focal length, expressed in the lensmaker's equation. For our exercise, we used the equation \( \frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \), assuming \( R_1 = -R_2 = R \). Solving this equation gives is a way to find the radius needed for the desired focal length, in this case, 7.04 mm.

Understanding radii of curvature offers insight into how lens shape influences its optical properties, such as focal length and bending of light.

Image Formation

Image formation in lenses, including the crystalline lens, is dictated by how they bend and focus light. For a lens, the principal focal point is where light rays converge to create a clear image. The position of the image depends on the object's distance from the lens and the focal length.

The thin lens formula \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \) helps determine the image position (\( d_i \)) given an object distance (\( d_o \)). In our exercise, placing an object 30 cm away and working with a given focal length allowed us to find the image's distance from the lens.

Recognizing whether an image is real or virtual, as well as its orientation, involves analyzing these calculations. A real image is formed when light actually converges at a point, resulting in an inverted image, as seen in our solution.

Thin Lens Formula

The thin lens formula is a fundamental equation used to relate object distance, image distance, and the lens's focal length. This simple yet powerful equation \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \) serves several purposes in both rudimentary and advanced optical problem-solving.

In practical applications, such as determining where an image will be focused, this formula allows us to solve for unknowns given certain initial conditions. In our exercise, it helped find the location of the image created by the eye lens, using a known object distance and focal length.

Understanding this formula equips you to predict basic lens behaviors, like image distance and magnification, and hence, is indispensable in optics and vision science.