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The focal length of the eyepiece of a certain microscope is 18.0 mm. The focal length of the objective is 8.00 mm. The distance between objective and eyepiece is 19.7 cm. The final image formed by the eyepiece is at infinity. Treat all lenses as thin. (a) What is the distance from the objective to the object being viewed? (b) What is the magnitude of the linear magnification produced by the objective? (c) What is the overall angular magnification of the microscope?

Short Answer

Expert verified
(a) 8.48 mm. (b) -21.1. (c) -293.3.

Step by step solution

01

Understand the Problem

We are given a microscope with the eyepiece and objective having focal lengths of 18.0 mm and 8.00 mm, respectively. The lenses are 19.7 cm apart, with the final image at infinity. We need to find the object distance for the objective, the linear magnification, and the microscope's angular magnification.
02

Relation for Lens Equation

We use the lens formula for the objective: \( \frac{1}{f_o} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f_o \) is the focal length of the objective (8.00 mm), \( d_o \) is the object distance, and \( d_i \) is the image distance. Since lenses are thin, \( d_i + d_e = 19.7 \text{ cm} \). Convert this to mm, giving 197 mm.
03

Find the Image Distance for Objective

Because the final image is at infinity, the image formed by the objective must be at the focal point of the eyepiece, \( d_e = 18 \text{ mm} \). Then, \( d_i = 197 \text{ mm} - 18 \text{ mm} = 179 \text{ mm}. \) Substitute into the lens equation to find \( d_o \).
04

Solve for Objective Object Distance

Using \( \frac{1}{8} = \frac{1}{d_o} + \frac{1}{179} \), solve for \( d_o \). Rearrange to get \( \frac{1}{d_o} = \frac{1}{8} - \frac{1}{179} \). Calculate to find \( d_o \approx 8.48 \text{ mm} \).
05

Linear Magnification of the Objective

The magnification \( m_o \) of the objective is given by \( m_o = -\frac{d_i}{d_o} \). Substitute \( d_i = 179 \text{ mm} \) and \( d_o = 8.48 \text{ mm} \). Calculate to find \( m_o \).
06

Overall Angular Magnification

The total magnification is the product of the objective's linear magnification and the eyepiece's angular magnification. The angular magnification of the eyepiece \( m_e = \frac{25 \, \text{cm}}{f_e} \), where \( 25 \, \text{cm} \) is the near point of human vision. Substitute \( f_e = 18 \text{ mm} = 1.8 \text{ cm} \). Calculate the final magnification \( m_{total} = m_o \times m_e \).
07

Final Calculations

Calculate \( m_o = -\frac{179}{8.48} \approx -21.1 \). Then \( m_e = \frac{25}{1.8} \approx 13.9 \), giving \( m_{total} = (-21.1) \times 13.9 \approx -293.29 \).

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Key Concepts

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

Focal Length
Focal length is a fundamental concept in optics that determines how strongly a lens converges or diverges light. It is the distance between the lens and its focus, where parallel light rays converge after passing through the lens.
  • A shorter focal length means the lens is more powerful and converges light quickly.
  • A longer focal length indicates a weaker lens that takes longer to bring light to a focus.
In the context of the provided exercise, the focal lengths play a crucial role in understanding how the microscope lenses function. The eyepiece has a focal length of 18.0 mm, while the objective's focal length is 8.00 mm. These measurements affect how magnification and image formation occur within the microscope.
The objective with a shorter focal length is capable of creating a highly magnified intermediate image, which the eyepiece further magnifies for the viewer.
Lens Formula
The lens formula is a key equation in optics, used to relate the object distance, image distance, and focal length of a lens. The standard form of this formula is:
\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]where:
  • \( f \) is the focal length of the lens.
  • \( d_o \) is the distance from the object to the lens.
  • \( d_i \) is the distance from the lens to the image.
In the original exercise, we use this formula to find the distance from the objective to the object being viewed. It's important to convert all measurements into consistent units, ensuring accurate calculations. For this problem, we derive that the image distance of the objective is effectively 179 mm, allowing us to calculate the object distance accurately as approximately 8.48 mm.
Magnification
Magnification in optics describes how much larger or smaller an image appears compared to the real object. It can be linear (for distances) or angular (for angles and visual perception).
Linear magnification for a lens, like the objective in a microscope, is given by:
  • \( m = -\frac{d_i}{d_o} \)
In this equation, \( m \) represents magnification, \( d_i \) is the image distance, and \( d_o \) is the object distance. For the microscope's objective with the calculated distances, it becomes clear the magnification is about \(-21.1\), indicating the image is inverted and magnified 21.1 times.
The total magnification of the microscope combines the eyepiece's angular magnification with the objective's linear magnification, providing the viewer with a greatly enhanced view of the object.
Thin Lens Equation
In optics, the thin lens equation is a simplified model that assumes lenses have negligible thickness. This model allows for easier calculations, particularly in educational contexts or basic optical devices.
By treating lenses as thin, their focal lengths can be effectively modeled with the lens formula. Thin lens assumptions simplify complex systems, like microscopes, into manageable data points. This aids in calculating crucial measures, like object and image distances, and magnifications.
Within the provided exercise, assuming thin lenses helps find key distances and magnifications with minimal computational effort, using the relations derived from the lens formula.
While this model is quite effective for most calculations, it's important to remember that real lenses may have limited thickness, affecting precision in professional applications.

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Most popular questions from this chapter

A thin lens with a focal length of 6.00 cm is used as a simple magnifier. (a) What angular magnification is obtainable with the lens if the object is at the focal point? (b) When an object is examined through the lens, how close can it be brought to the lens? Assume that the image viewed by the eye is at the near point, 25.0 cm from the eye, and that the lens is very close to the eye.

A spherical, concave shaving mirror has a radius of curvature of 32.0 cm. (a) What is the magnification of a person's face when it is 12.0 cm to the left of the vertex of the mirror? (b) Where is the image? Is the image real or virtual? (c) Draw a principal-ray diagram showing the formation of the image.

To determine whether a frog can judge distance by means of the amount its lens must move to focus on an object, researchers covered one eye with an opaque material. An insect was placed in front of the frog, and the distance that the frog snapped its tongue out to catch the insect was measured with high-speed video. The experiment was repeated with a contact lens over the eye to determine whether the frog could correctly judge the distance under these conditions. If such an experiment is performed twice, once with a lens of power -9-D and once with a lens of power -15-D, in which case does the frog have to focus at a shorter distance, and why? (a) With the -9-D lens; because the lenses are diverging, the lens with the longer focal length creates an image that is closer to the frog. (b) With the -15-D lens; because the lenses are diverging, the lens with the shorter focal length creates an image that is closer to the frog. (c) With the -9-D lens; because the lenses are converging, the lens with the longer focal length creates a larger real image. (d) With the -15-D lens; because the lenses are converging, the lens with the shorter focal length creates a larger real image.

A mirror on the passenger side of your car is convex and has a radius of curvature with magnitude 18.0 cm. (a) Another car is behind your car, 9.00 m from the mirror, and this car is viewed in the mirror by your passenger. If this car is 1.5 m tall, what is the height of the image? (b) The mirror has a warning attached that objects viewed in it are closer than they appear. Why is this so?

A converging lens with a focal length of 12.0 cm forms a virtual image 8.00 mm tall, 17.0 cm to the right of the lens. Determine the position and size of the object. Is the image erect or inverted? Are the object and image on the same side or opposite sides of the lens? Draw a principal-ray diagram for this situation.

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