Problem 24 - University Physics with Modern Physics

Question

The left end of a long glass rod 8.00 cm in diameter, with an index of refraction of 1.60, is ground and polished to a convex hemispherical surface with a radius of 4.00 cm. An object in the form of an arrow 1.50 mm tall, at right angles to the axis of the rod, is located on the axis 24.0 cm to the left of the vertex of the convex surface. Find the position and height of the image of the arrow formed by paraxial rays incident on the convex surface. Is the image erect or inverted?

Step-by-Step Solution

Answer

Image is 8.35 cm to the right, height is 0.5205 mm, inverted.

Step 1: Identify the Given Parameters

We have a glass rod with a diameter of 8.00 cm, which means its radius is 4.00 cm. The index of refraction \( n = 1.60 \). The object (an arrow) is 1.50 mm tall and located at 24.0 cm from the vertex of the convex surface. The convex surface has a radius of 4.00 cm.

Step 2: Use the Lensmaker's Equation

To find the image position, use the lensmaker's equation for a curved refracting surface: \( \frac{n_2}{s'} - \frac{n_1}{s} = \frac{n_2 - n_1}{R} \), where \( n_1 = 1.00 \) (air), \( n_2 = 1.60 \), \( s = 24.0 \; \text{cm} \) (object distance), \( R = 4.00 \; \text{cm} \) (radius of curvature), and \( s' \) is the image distance, which we need to find.

Step 3: Calculate the Image Distance

Substitute the known values into the equation: \[ \frac{1.60}{s'} - \frac{1.00}{24.0} = \frac{1.60 - 1.00}{4.00} \]Simplify and solve for \( s' \): \[ \frac{1.60}{s'} - \frac{1}{24.0} = 0.15 \]\[ \frac{1.60}{s'} = 0.15 + \frac{1}{24.0} \]\[ \frac{1.60}{s'} = 0.15 + 0.0417 = 0.1917 \]\[ s' = \frac{1.60}{0.1917} \approx 8.35 \text{ cm} \]The image is formed 8.35 cm to the right of the vertex of the convex surface.

Step 4: Determine the Magnification

The magnification \( m \) is given by \(-\frac{s'}{s} \). We have \( s' = 8.35 \text{ cm} \) and \( s = 24.0 \text{ cm} \), thus:\[ m = -\frac{8.35}{24.0} = -0.347 \]

Step 5: Calculate the Image Height

Use the magnification to find the image height \(h'\) given \( h = 1.50 \text{ mm} \):\[ h' = m \times h = -0.347 \times 1.50 \text{ mm} = -0.5205 \text{ mm} \]The negative sign indicates that the image is inverted.

Step 6: Conclusion

The image of the arrow is located 8.35 cm to the right of the convex surface. It has a height of 0.5205 mm and is inverted.

Key Concepts

Refractive Index

The refractive index is a fundamental concept in optics that describes how light propagates through different media. It is denoted by the letter "n" and is defined as the ratio of the speed of light in vacuum to the speed of light in the medium. A higher refractive index indicates that light travels more slowly in that medium. For example, the refractive index of air is approximately 1, while the given glass rod in our exercise has a refractive index of 1.60. This means light travels slower in the glass compared to air.

Understanding the refractive index is crucial because it affects how light bends or refracts when it enters a new medium. Systems like glasses, lenses, and other optical devices rely on this bending property for effective image formation. When dealing with lenses or curved glass surfaces, such as in our exercise, knowing the refractive index allows us to predict how images will form when light passes through the material.

Refractive indices not only affect speed but also the direction of light rays. This can be calculated using Snell's Law, but for lenses and curved surfaces, as in the given glass rod scenario, we utilize specific equations like the lensmaker's equation to find image positions and other characteristics.

Lensmaker's Equation

The lensmaker's equation is a critical formula in optics for determining the focal length of a lens, or in this scenario, the image position formed by a refractive surface, which is similar in application. This equation helps us understand how the curvature of a surface and the refractive index of a material influence the convergence or divergence of light beams.

The lensmaker's equation used in our context of a spherical surface is: \[ \frac{n_2}{s'} - \frac{n_1}{s} = \frac{n_2 - n_1}{R} \]where:

• \( n_1 \) and \( n_2 \) are the refractive indices of the original and new medium, respectively.
• \( s \) is the object distance from the vertex of the lens (or surface), and \( s' \) is the image distance, which is what we solve for.
• \( R \) is the radius of curvature of the lens surface.

In our exercise, this equation helps find where the image of an object lies after light has passed through the polished glass rod. By substituting the given values into the equation, we determine the position of the image formed by the rod.

The understanding gained from the lensmaker's equation forms the foundation for many designing improvements in optical devices, providing a detailed prediction of how changes in curvature and materials affect image formation.

Optical Image Formation

Optical image formation is a key aspect of understanding how light interacts with lenses and curved surfaces to create images. In the realm of optics, this involves the bending of light rays upon entering a medium with a different refractive index, guided by principles like the lensmaker's equation.

When light rays pass through a curved surface, such as the convex surface of the glass rod in our exercise, they are refracted and converge at a specific point to form an image. The position and nature of this image—whether it is real or virtual, upright or inverted—depend on factors such as:

• The refractive index of the material.
• The curvature of the surface involved.
• The distance of the object from the surface.

In our scenario, an optical image is formed 8.35 cm on the other side of the convex surface after applying these principles. Additionally, using magnification, we can determine the size and orientation of the image. With a negative magnification of \(-0.347\), it is clear that the image is inverted and smaller than the original object.

By comprehending these optical principles, students can effectively predict and analyze how images are formed in varied scenarios, which is essential for the study of optics and the development of visual technologies.