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One end of a long glass rod \((n=1.50)\) is formed into a convex surface with a radius of curvature of \(6.00 \mathrm{cm} .\) An object is located in air along the axis of the rod. Find the image positions corresponding to object distances of (a) \(20.0 \mathrm{cm},(\mathrm{b}) 10.0 \mathrm{cm},\) and \((\mathrm{c}) 3.00 \mathrm{cm}\) from the end of the rod.

Short Answer

Expert verified
For object distances of 20.0 cm, 10.0 cm and 3.00 cm, the corresponding image distances can be found by solving the refractive mirror equation for each case individually after substituting the given values.

Step by step solution

01

- Understand the Mirror Equation for Refraction

We need to use the lens-makers' equation for refraction at a spherical surface, which is given by \(\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{r}\), where \(n_2\) is the refractive index of glass (1.50), \(n_1\) is the refractive index of air (1.00), \(v\) is the image distance, \(u\) is the object distance, and \(r\) is the radius of curvature of the convex surface. Since the surface is convex and the object is in air, the radius of curvature will be positive, so \(r = +6.00\, \mathrm{cm}\).
02

- Case (a) Find Image Distance for Object Distance of 20.0 cm

For the object distance of \(20.0\,\mathrm{cm}\), we plug the values into the mirror equation: \(\frac{1.50}{v} - \frac{1.00}{-20.0} = \frac{1.50 - 1.00}{6.00}\).Then solve for \(v\). Note that the object distance is negative in the mirror equation because the light is coming from the object.
03

- Solve for Image Distance in Case (a)

Rearrange and solve the equation for \(v\): \(\frac{1.50}{v} = \frac{1.00}{-20.0} + \frac{0.50}{6.00}\).By doing the math, we find \(v\) for case (a).
04

- Case (b) Find Image Distance for Object Distance of 10.0 cm

For the object distance of \(10.0\,\mathrm{cm}\), repeat the process: \(\frac{1.50}{v} - \frac{1.00}{-10.0} = \frac{1.50 - 1.00}{6.00}\).Solve for \(v\) for this case.
05

- Case (c) Find Image Distance for Object Distance of 3.00 cm

For the object distance of \(3.00\,\mathrm{cm}\), use the equation again: \(\frac{1.50}{v} - \frac{1.00}{-3.00} = \frac{1.50 - 1.00}{6.00}\).Find \(v\) for this last case.
06

- Calculate the Numerical Values for Each Case

Carry out the calculations for each object distance to determine the image positions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Lens-Makers' Equation
The lens-makers' equation is a foundational concept in optics, particularly when it comes to understanding how lenses form images. This formula relates the physical characteristics of a lens or a spherical surface with the way it bends, or refracts, light. When an object is placed in front of a lens or spherical surface, such as the end of a glass rod, the equation
\[\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{r}\]
allows us to calculate where the image formed by the surface will be located relative to the lens. Here, \(n_1\) and \(n_2\) are the refractive indices of the media on either side of the surface, \(u\) is the object distance from the lens, \(v\) is the image distance, and \(r\) is the radius of curvature of the lens. By plugging in the given values, we can find out precisely where the image will form for various object distances. This is crucial for lens design, enabling us to tailor optical devices for specific applications.
Radius of Curvature and Its Impact on Optics
The radius of curvature is a term that may sound complex but is really quite simple—it's the radius of the sphere from which a lens or a mirror surface is a segment. This measurement plays a crucial role in the behavior of refracting or reflecting surfaces. A positive radius of curvature indicates a surface that bulges outwards, like a convex lens or mirror, while a negative value would imply an inward dip, as seen in concave lenses or mirrors.
In our exercise, the radius of curvature is positive, indicating a convex refractive surface. This convexity affects how light rays converge or diverge when passing through the surface. For optical calculations, this value feeds directly into the lens-makers' equation, helping to determine where an image will form in relation to the surface. A larger radius means a gentler curve, which generally results in a less dramatic bending of light, whereas a smaller radius leads to sharper curvature, impacting how and where images are formed.
Calculating Image Distance in Optics
Calculation of image distance is a common task when dealing with optical systems. By using the lens-makers' equation, we can determine the position of the image created by a lens or refractive surface from a given object distance. In practical terms, understanding image distance is essential for applications like focusing a camera, designing glasses, or setting up an optical experiment.

Image Distance Calculation Steps

  • Identify the refractive indices of the media involved (\(n_2\) for the lens material, \(n_1\) for the surrounding medium).
  • Apply the known object distance (\(u\)) and the radius of curvature (\(r\)) of the lens.
  • Rearrange the lens-makers' equation to solve for \(v\), the image distance.
  • Perform the mathematical operations to find the value of \(v\).

Once these steps are completed for the provided object distances, we can predict where an image will appear in space, allowing for precise control over optical systems.
Refractive Index and Its Role in Optics
Refractive index, symbolized by 'n', is a measure of how much a ray of light bends, or refracts, as it passes from one medium into another. Every transparent medium has its own refractive index, which determines the speed at which light travels through it. Air, for example, has a refractive index close to 1, while glass and water have higher indices, indicating they slow down light more and bend its path to a greater extent.

Significance of Refractive Index in Calculations

  • The difference in refractive indices between two media controls the bending of light at their interface.
  • It's a crucial part of the lens-makers' equation, directly affecting where an image is formed.
  • Gaining an accurate measurement of refractive index is vital for constructing optical instruments and analyzing their behavior.

In our example, the refractive index of air (1) is much lower than that of the glass rod (1.50), indicating that as light enters the glass from the air, it significantly bends towards the normal. This bend is what allows the convex surface to focus light and create images at specific distances, fundamentally impacting how the image is perceived.

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Most popular questions from this chapter

A compound microscope has an objective of focal length 0.300 \(\mathrm{cm}\) and an eyepiece of focal length \(2.50 \mathrm{cm} .\) If an object is 3.40 \(\mathrm{mm}\) from the objective, what is the magnification? (Suggestion: Use the lens equation for the objective.)

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A cataract-impaired lens in an eye may be surgically removed and replaced by a manufactured lens. The focal length required for the new lens is determined by the lens-to-retina distance, which is measured by a sonar-like device, and by the requirement that the implant provide for correct distant vision. (a) Assuming the distance from lens to retina is \(22.4 \mathrm{mm},\) calculate the power of the implanted lens in diopters. (b) Because no accommodation occurs and the implant allows for correct distant vision, a corrective lens for close work or reading must be used. Assume a reading distance of 33.0 \(\mathrm{cm}\) and calculate the power of the lens in the reading glasses.

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