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Chapter 34: Problem 89

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 rod's length measured between the vertices of the two hemispherical surfaces is calculated using optical formulas and found to be consistent with the placement of the final image.

Step by step solution

01

Understanding the Sign Convention

In optics, we use certain sign conventions. For convex surfaces, the radius of curvature \( R \) is positive if the surface is facing towards the object. We also consider that distances measured in the same direction as the incident light are positive.
02

Calculate Image Distance for Left Surface

For the left hemispherical surface, use the lens maker's formula: \[\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}\]where \( n_1 = 1 \) (air), \( n_2 = 1.55 \), \( u = -25.0 \text{ cm} \) (object distance; negative because the object is on the same side as the light), and \( R = 6.00 \text{ cm} \). Substituting gives:\[\frac{1.55}{v} - \frac{1}{-25} = \frac{1.55 - 1}{6}\]Solve for \( v \), which is the image distance inside the rod after the first surface.
03

Solve for Image Distance inside Rod

Rearrange and solve:\[\frac{1.55}{v} + \frac{1}{25} = \frac{0.55}{6}\]Approximate to:\[v \approx 20.5 \text{ cm}\]This is the image created by the left surface inside the rod.
04

Calculate Image Distance for Right Surface

Use similar reasoning and the above formula to find the image distance to the right surface.Here, the object distance \( u \) for the right hemispherical surface will be \( 20.5 \text{ cm} \) (positive since it is measured from inside the rod towards infinity). Use the radius \( R = -6.00 \text{ cm} \) because it faces away from the internal medium.Substitute:\[\frac{1}{v_r} - \frac{1.55}{20.5} = \frac{1 - 1.55}{-6}\]Solve for \( v_r \) to find the final image's distance from the pole of the second surface.
05

Calculate Length of the Rod

The final image distance \( v' = 65.0 \text{ cm} \) to the right of the second surface given previously should equal the calculated \( v_r \). The length \( L \) of the rod can be found by calculating the spacing between these two images plus the length obtained in Step 3.Thus:\[L = v_r - 20.5\],solving which gives the length of the rod.

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Key Concepts

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

Refractive Index
The refractive index is a fundamental concept in geometrical optics that measures how much light slows down as it passes through a medium. It is represented by the symbol "n" and is defined as the ratio of the speed of light in a vacuum to the speed of light in the medium. For instance, the refractive index of glass is often around 1.5, indicating that light travels 1.5 times slower in glass than in a vacuum.

An important point to remember is that the refractive index influences how much a light ray bends, or "refracts," when it enters a new medium at an angle. The greater the difference in refractive indices between two media, the more the light will bend at the interface. This concept is crucial when designing lenses and optical instruments.

In our specific problem, the glass rod has a refractive index of 1.55, indicating how light will behave as it passes through the rod's material.
Lens Maker's Formula
The Lens Maker's Formula is a key equation in optics that relates the focal length of a lens to its shape and the refractive index of its material. The formula is expressed as:\[\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}\]In this equation:
  • \(n_1\) and \(n_2\) are the refractive indices of the initial and final media, respectively.
  • \(v\) is the image distance.
  • \(u\) is the object distance.
  • \(R\) is the radius of curvature of the lens surfaces.
Depending on the curvature of the lens surfaces and the refractive index difference, this formula helps in predicting how the lens will form images.

For the exercise problem, this formula is used twice, for both ends of the rod which are shaped like hemispherical surfaces. By applying it, we can determine the positions of the images formed by these surfaces. This enables us to eventually calculate the length of the rod that matches the image formed outside of it.
Image Distance Calculation
Image distance calculation is an essential aspect of understanding how lenses and curved surfaces create images. In optics, we often derive image distances using formulas such as the Lens Maker's Formula mentioned earlier. Once we have applied the formula, solving for the variable \(v\) or \(v_r\) gives us the position where the image is formed.
The problem involves calculating two critical image distances. Firstly, for the left surface of the rod: after substituting values in the formula, the image distance within the rod was found to be approximately 20.5 cm. This means that from the point of refraction at the left surface, the light focuses 20.5 cm inside the rod.
Secondly, for the right surface of the rod: using a similar process with adjusted values (like the object distance being the internal image position and an opposite radius sign), the final image distance aligns with the given distance of 65.0 cm from the second surface.
Together, these calculations allow us to find the length of the rod as the distance between these internal image points plus any given external distances.

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Most popular questions from this chapter

A light bulb is 3.00 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?

Where must you place an object in front of a concave mirror with radius \(R\) so that the image is erect and 2\(\frac{1}{2}\) times the size of the object? Where is the image?

The left end of a long glass rod 6.00 \(\mathrm{cm}\) in diameter has a convex hemispherical surface 3.00 \(\mathrm{cm}\) in radius. The refractive index of the glass is 1.60. Determine the position of the image if an object is placed in air on the axis of the rod at the following distances to the left of the vertex of the curved end: (a) infinitely far, (b) \(12.0 \mathrm{cm} ;(c) 2.00 \mathrm{cm} .\)

What Is the Smallest Thing We Can See? The smallest object we can resolve with our eye is limited by the size of the light receptor cells in the retina. In order for us to distinguish any detail in an object, its image cannot be any smaller than a single retinal cell. Although the size depends on the type of cell (rod or cone), a diameter of a few microns \((\mu \mathrm{m})\) is typical near the center of the eye. We shall model the eye as a sphere 2.50 \(\mathrm{cm}\) in diameter with a single thin lens at the front and the retina at the rear, with light receptor cells 5.0\(\mu \mathrm{m}\) in diameter. (a) What is the smallest object you can resolve at a near point of 25 \(\mathrm{cm}\) ? (b) What angle is subtended by this object at the eye? Express your answer in units of minutes \(\left(1^{\circ}=60 \mathrm{min}\right),\) and compare it with the typical experimental value of about 1.0 min. (Note: There are other limitations, such as the bending of light as it passes through the pupil, but we shall ignore them here.)

Combination of Lenses II. Two thin lenses with a focal length of magnitude \(12.0 \mathrm{cm},\) the first diverging and the second converging, are located 9.00 \(\mathrm{cm}\) apart. An object 2.50 \(\mathrm{mm}\) tall is placed 20.0 \(\mathrm{cm}\) to the left of the first (diverging) lens. (a) How far from this first lens is the final image formed? (b) Is the final image real or virtual? (c) What is the height of the final image? Is it erect or inverted? (Hint: See the preceding two problems.)
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