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Chapter 34: Problem 88

A glass rod with a refractive index of 1.55 is ground and polished at both ends to hemispherical surfaces with radii of 6.00 \(\mathrm{cm}\) . When an object is placed on the axis of the rod, 25.0 \(\mathrm{cm}\) to the left of the left-hand end, the final image is formed 65.0 \(\mathrm{cm}\) to the right of the right-hand end. What is the length of the rod measured between the vertices of the two hemispherical surfaces?

Short Answer

Expert verified
The length of the rod is 29.92 cm.

Step by step solution

01

Understand the Problem Setup

The glass rod has two hemispherical surfaces with each having a radius of 6.00 cm. The object distance from the left end is 25.0 cm and the final image distance from the right end is 65.0 cm. We need to determine the length of the rod, which means determining the distance between the vertices of the two spheres.
02

Apply the Lensmaker's Formula

Since each end of the rod acts like a lens, we can consider each to have a focal length. Since the surface is hemispherical, the focal length for each surface can be calculated using the lensmaker's formula: \[ f = \frac{R}{2(n-1)} \] where \( R = 6.00 \, \mathrm{cm} \) and \( n = 1.55 \). Thus, we find that:\[ f = \frac{6.00}{2(1.55 - 1)} = 10.91 \, \mathrm{cm} \]
03

Determine Object Distance for Each Lens

For the left hemisphere, the object is 25.0 cm to the left, while the focal length is 10.91 cm. Using the lens formula \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \), where \( u = -25.0 \, \mathrm{cm} \), we solve for \( v \) (image distance from the first lens):\[ \frac{1}{v} = \frac{1}{10.91} + \frac{1}{25.0} \]\[ v = 17.19 \, \mathrm{cm} \] (to the right of the left end).
04

Calculate Image Distance from Second Surface

For the second surface, the image formed by the first surface becomes the object for the second surface. The object distance from the second hemisphere is calculated as \( d - 17.19 \, \mathrm{cm}\), where \( d \) is the length of the rod. If \( f = 10.91 \, \mathrm{cm} \), final image distance after both surfaces is 65.0 cm to the right of the second end:\[ \frac{1}{v_2} - \frac{1}{(d - 17.19)} = \frac{1}{10.91} \]\[ \frac{1}{65.0} = \frac{1}{10.91} - \frac{1}{(d - 17.19)} \]
05

Solve for the Rod Length

Rearranging the equation gives us: \[ d - 17.19 = \left( \frac{1}{10.91} - \frac{1}{65.0} \right)^{-1} \]Solving, we find:\[ d - 17.19 = 12.73 \, \mathrm{cm} \]Thus:\[ d = 29.92 \, \mathrm{cm} \]
06

Conclusion of Calculation

The length of the rod measured between the vertices of the two hemispherical surfaces is 29.92 cm. This length keeps consistent with the object and image positions relative to each end of the rod with their refractive properties.

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Key Concepts

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

Lensmaker's Formula
Understanding the properties of lenses is essential when studying geometrical optics. One of the fundamental tools in this field is the Lensmaker's Formula. This formula helps us calculate the focal length of a lens based on its physical attributes and the material from which it is made. The Lensmaker's Formula is given by:\[ f = \frac{R}{2(n-1)}\]where:
  • \( f \) is the focal length of the lens.
  • \( R \) is the radius of curvature of the lens surface.
  • \( n \) is the refractive index of the lens material.
In our exercise, each hemispherical end of the glass rod is considered a lens. With a refractive index of 1.55 and a radius of 6.00 cm, we can use this formula to find the focal length as 10.91 cm. This calculation is crucial as it allows us to determine how light will behave when it passes through the lens surfaces of the rod, helping to solve complex problems like evaluating the rod's length by using the principles of optics.
Refractive Index
The refractive index is an important concept in understanding how light behaves as it transitions between different media. It measures how much the speed of light is reduced inside a medium compared to vacuum.In mathematical terms, the refractive index \( n \) is defined as:
  • \( n = \frac{c}{v} \)
where:
  • \( c \) is the speed of light in a vacuum.
  • \( v \) is the speed of light in the medium.
When an object like our glass rod has a refractive index (1.55 in this exercise), it indicates how much the light will bend when it passes through. This value is greater than 1, showing that light travels slower in the glass than in vacuum, which affects how images are formed.Understanding the refractive index helps us predict how lenses within the rod will bend light and where the resulting images will form. It's a foundational concept for comprehending how lenses are designed and help in solving problems involving light and optics.
Image Formation
Image formation is a key outcome of light interacting with lenses and surfaces. The way light converges or diverges after passing through a lens determines where and how an image forms. Understanding image formation involves applying the lens formula:\[\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\]where:
  • \( v \) is the image distance from the lens.
  • \( u \) is the object distance from the lens.
  • \( f \) is the focal length of the lens.
In the problem, an object positioned 25.0 cm to the left of the rod is processed by the lens-like ends of the rod to form an image. The initial image forms 17.19 cm to the right of the left end, becoming the object for the second hemispherical lens, eventually giving a final image 65.0 cm to the right of the second end of the rod.By using the lens formula and the concept of image formation, we can solve for unknown distances, like calculating the rod length in our exercise, and understand the overall process of how and where images are made. This critical insight into optics enables us to design lenses and optical systems that perform specific functions.

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Most popular questions from this chapter

You want to view an insect 2.00 \(\mathrm{mm}\) in length through a magnifier. If the insect is to be at the focal point of the magnifier. what focal length will give the image of the insect an angular size of 0.025 radian?

An object to the left of a lens is imaged by the lens on a screen 30.0 \(\mathrm{cm}\) to the right of the lens. When the lens is moved 4.00 \(\mathrm{cm}\) to the right, the screen must be moved 4.00 \(\mathrm{cm}\) to the left to refocus the image. Determine the focal length of the lens.

You wish to project the image of a slide on a screen 9.00 \(\mathrm{m}\) from the lens of a slide projector. (a) If the slide is placed 15.0 \(\mathrm{cm}\) from the lens, what focal length lens is required? (b) If the dimensions of the picture on a \(35-\mathrm{mm}\) color slide are \(24 \mathrm{mm} \times 36 \mathrm{mm},\) what is the minimum size of the projector screen required to accommodate the image?

A light bulb is 4.00 \(\mathrm{m}\) from a wall. You are to use a concave mirror to project an image of the bulb on the wall, with the image 2.25 times the size of the object. How far should the mirror be from the wall? What should its radius of curvature be?

Focus of the Eye. The comea of the eye has a radius of curvature of approximately \(0.50 \mathrm{cm},\) and the aqueous humor behind it has an index of refraction of \(1.35 .\) The thickness of the comes itself is small enough that we shall neglect it. The depth of a typical human eye is around 25 \(\mathrm{mm}\) (a) What would have to be the radius of curvature of the cornea so that it alone would focus the image of a distant mountain on the retina, which is at the back of the eye opposite the cornea? (b) If the cornea focused the mountain correcly on the retina as described in part (a), would it also focus the text from a computer screen on the retina if that screen were 25 \(\mathrm{cm}\) in front of the eye? If not, where would it focus that text in front of or behind the retina? (c) Given that the cornea has a radius of curvature of about 5.0 \(\mathrm{mm}\) , where does it actually focus the mountain? Is this in front of or behind the retina? Does this help you see why the eye needs help from a lens to complete the task of focusing?
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