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Chapter 34: Problem 20

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) \(12.0 \mathrm{cm} ;(c) 2.00 \mathrm{cm} .\)

Short Answer

Expert verified
(a) 8.00 cm inside, (b) 13.71 cm inside, (c) 5.33 cm outside.

Step by step solution

01

Understanding the Lens Maker's Formula

The lens maker's formula is used to determine the position of the image formed by spherical surfaces. It 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 the glass (1.60), \(n_1\) is the refractive index of air (approximately 1.00), \(u\) is the object distance, \(v\) is the image distance, and \(R\) is the radius of curvature of the hemispherical surface.
02

Solving for an Object at Infinity

When an object is at infinity, the incoming rays are parallel and the image is formed at the focal point of the lens. In this case:\[u = \infty\]Thus, \(\frac{n_1}{u} = 0\). Substituting in the lens maker's formula, we get:\[\frac{n_2}{v} = \frac{n_2 - n_1}{R}\]Substitute \(n_2 = 1.60\), \(n_1 = 1.00\), and \(R = 3.00\, \mathrm{cm}\) to find \(v\):\[v = \frac{n_2 \cdot R}{n_2 - n_1} = \frac{1.60 \times 3.00}{1.60 - 1.00} = 8.00\, \mathrm{cm}\]The image is formed 8.00 cm inside the glass rod.
03

Solving for an Object at 12.0 cm

For an object at a distance of 12.0 cm:\[u = -12.0\, \mathrm{cm}\] (negative since object is on the left)Using the lens maker's formula:\[\frac{1.60}{v} - \frac{1.00}{(-12.0)} = \frac{0.60}{3.00}\]Simplify and solve for \(v\):\[\frac{1.60}{v} + \frac{1}{12} = \frac{0.60}{3}\]\[\frac{1.60}{v} = 0.2 - 0.0833 = 0.1167\]\[v = \frac{1.60}{0.1167} \approx 13.71\, \mathrm{cm}\]The image is formed 13.71 cm inside the glass rod.
04

Solving for an Object at 2.00 cm

For an object at a distance of 2.00 cm:\[u = -2.00\, \mathrm{cm}\]Using the lens maker's formula:\[\frac{1.60}{v} - \frac{1.00}{(-2.00)} = \frac{0.60}{3.00}\]Simplify and solve for \(v\):\[\frac{1.60}{v} + 0.5 = 0.2\]\[\frac{1.60}{v} = 0.2 - 0.5 = -0.3\]\[v = \frac{1.60}{-0.3} = -5.33\, \mathrm{cm}\]The image is formed 5.33 cm to the left of the vertex, outside the glass 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 optics that describes how light propagates through a medium. It is denoted 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 the given problem, we have:
  • The refractive index of glass (\(n_2\)) is 1.60.
  • The refractive index of air (\(n_1\)) is approximately 1.00.
This indicates that light travels slower in glass than in air. The refractive index affects how much light bends when entering a new medium, which is crucial in lens calculations. A higher refractive index means the light bends more. In the lens maker's formula, refractive index values are used to determine how a lens focuses light. This is essential for calculating image distances in optical devices.
Hemispherical Surface
A hemispherical surface, in this context, refers to the round, convex end of the glass rod described in the problem. This part of the lens has a specific role in the way it refracts light beams passing through it. Here are some key characteristics:
  • The radius of curvature (\(R\)) of the hemispherical surface is 3.00 cm. It represents the radius of the sphere from which this hemispherical lens surface is a part.
  • Its convex shape means it curves outward, which affects how light converges or diverges upon entering this surface from another medium.
When dealing with such surfaces, understanding the shape and radius of the surface is crucial, as they influence the focus and formation of images through refraction. The lens maker's formula incorporates this radius to find where light will gather to form an image.
Image Distance Calculation
Image distance calculation involves determining where an image will form relative to a lens or curved surface. The main formula used is the lens maker's formula:\[\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}\]Here:
  • \(v\) is the image distance, the distance from the lens to where the image is formed.
  • \(u\) is the object distance, the distance from the object to the lens.
  • Using positive and negative signs is crucial. In this problem, a negative object distance indicates the object is placed to the left of the lens.
  • As demonstrated in the solutions, setting different object distances shows their impact on image formation within or outside the lens material.
The approach involves substituting known values into the formula to solve for the unknown variable, \(v\). This allows us to see whether the image is real and on the same side as the outgoing light or virtual and needing further understanding of the setup to be visualized.

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Most popular questions from this chapter

The Lens of the Eye. 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 \(\mathrm{mm}\) , 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 \(\mathrm{cm}\) tall is placed 30.0 \(\mathrm{cm}\) 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.)

Curvature of the Cornea. In a simplified model of the human eye, the aqueous and vitreous humors and the lens all have a refractive index of 1.40 , and all the refraction occurs at the cornea, whose vertex is 2.60 \(\mathrm{cm}\) from the retina. What should be the radius of curvature of the cornea such that the image of an object 40.0 \(\mathrm{cm}\) from the cornea's vertex is focused on the retina?

Focal Length of a Zoom Lens. Figure P34.113 shows a simple version of a zoom lens. The converging lens has focal length \(f_{1},\) and the diverging lens has focal length \(f_{2}=-\left|f_{2}\right|\) The two lenses are separated by a variable distance \(d\) that is always less than \(f_{1} .\) Also, the magnitude of the focal length of the diverging lens satisfies the inequality \(\left|f_{2}\right|>\left(f_{1}-d\right) .\) To determine the effective focal length of the combination lens, consider a bundle of parallel rays of radius \(r_{0}\) entering the converging lens. (a) Show that the radius of the ray bundle decreases to \(r_{0}^{\prime}=r_{0}\left(f_{1}-d\right) / f_{1}\) at the point that it enters the diverging lens. (b) Show that the final image \(I^{\prime}\) is formed a distance \(s_{2}^{\prime}=\left|f_{2}\right|\left(f_{1}-d\right) /\left(\left|f_{2}\right|-f_{1}+d\right)\) to the right of the diverging lens. (c) If the rays that emerge from the diverging lens and reach the final image point are extended backward to the left of the diverging lens, they will eventually expand to the original radius \(r_{0}\) at some point \(Q .\) The distance from the final image \(I^{\prime}\) to the point \(Q\) is the effective focal length \(f\) of the lens combination; if the combination were replaced by a single lens of focal length \(f\) placed at \(Q,\) parallel rays would still be brought to a focus at \(I^{\prime}\) . Show that the effective focal length is given by \(f=f_{1}\left|f_{2}\right| /\left(\left|f_{2}\right|-f_{1}+d\right) .\) (d) If \(f_{1}=12.0 \mathrm{cm}\) \(f_{2}=-18.0 \mathrm{cm},\) and the separation \(d\) is adjustable between 0 and \(4.0 \mathrm{cm},\) find the maximum and minimum focal lengths of the combination. What value of \(d\) gives \(f=30.0 \mathrm{cm} ?\)

A telescope is constructed from two lenses with focal lengths of 95.0 \(\mathrm{cm}\) and \(15.0 \mathrm{cm},\) the 95.0 -cm lens being used as the objective. Both the object being viewed and the final image are at infinity. (a) Find the angular magnification for the telescope. (b) Find the height of the image formed by the objective of a building 60.0 \(\mathrm{m}\) tall, 3.00 \(\mathrm{km}\) away. (c) What is the angular size of the final image as viewed by an eye very close to the eyepiece?

cp calc You are in your car driving on a highway at 25 \(\mathrm{m} / \mathrm{s}\) when you glance in the passenger-side mirror (a convex mirror with radius of curvature 150 \(\mathrm{cm}\) ) and notice a truck approaching. If the image of the truck is approaching the vertex of the mirror at a speed of 1.9 \(\mathrm{m} / \mathrm{s}\) when the truck is 2.0 \(\mathrm{m}\) from the mirror, what is the speed of the truck relative to the highway?
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