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Chapter 25: Problem 29

A compound microscope has a distance of \(15 \mathrm{~cm}\) between lenses and an objective with a focal length of \(8.0 \mathrm{~mm} .\) What power should the eyepiece have to give a total magnification of \(-360 \times ?\)

Short Answer

Expert verified
The eyepiece should have a power of approximately \(-80.8 \mathrm{~D}\).

Step by step solution

01

Understand the total magnification equation

The total magnification \( M_t \) of a compound microscope is given by the product of the magnifications of the objective lens \( M_o \) and the eyepiece \( M_e \): \( M_t = M_o \times M_e \).
02

Calculate the magnification of the objective lens

The magnification of the objective lens \( M_o \) is given by the formula \( M_o = \frac{d}{f_o} \), where \( d \) is the tube length (distance between the lenses) and \( f_o \) is the focal length of the objective lens. Here, \( d = 15 \mathrm{~cm} = 150 \mathrm{~mm} \) and \( f_o = 8.0 \mathrm{~mm} \). Thus, \( M_o = \frac{150}{8} = 18.75 \times \).
03

Calculate the required magnification of the eyepiece

We know that the total magnification \( M_t = -360 \times \). Using the equation \( M_t = M_o \times M_e \), solve for the eyepiece magnification \( M_e \): \( -360 = 18.75 \times M_e \). Therefore, \( M_e = \frac{-360}{18.75} \approx -19.2 \times \).
04

Calculate the power of the eyepiece

The power of a lens \( P \), or diopter, is the reciprocal of its focal length in meters: \( P = \frac{1}{f_e} \). Since the magnification of the eyepiece \( M_e \) is given by \( M_e = 1 + \frac{250}{f_e} \) (assuming normal near point as 25 cm or 250 mm), equate \( 1 + \frac{250}{f_e} = -19.2 \). Solving for \( f_e \): \( \frac{250}{f_e} = -20.2 \). Thus, \( f_e = \frac{250}{-20.2} \approx -12.38 \mathrm{~mm} \). Finally, the power is \( P = \frac{1000}{-12.38} \approx -80.8 \mathrm{~D} \).

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Key Concepts

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

Magnification
In a compound microscope, magnification is a critical concept as it defines how much larger the image appears compared to the actual object. The total magnification of the microscope is determined by multiplying the magnification of the objective lens by the magnification of the eyepiece. The formula is:
  • Mathematically: \( M_t = M_o \times M_e \) where \( M_t \) is the total magnification, \( M_o \) is the magnification of the objective, and \( M_e \) is the magnification of the eyepiece.
  • The total magnification tells us how many times larger the image is from the original object.
Understanding this can help in calculating the desired level of clarity needed for different types of observations.
Focal Length
The focal length is essential in determining how lenses in a compound microscope affect image magnification. The focal length is the distance between the lens and the point where it converges light into a focus.
  • For the objective lens, the focal length \( f_o \) is usually short. This short focal length means higher magnification because the lens bends light rays more sharply.
  • The formula to find the magnification of the objective lens is \( M_o = \frac{d}{f_o} \) where \( d \) is the distance between the lenses, also known as the tube length.
  • Shorter focal lengths lead to greater magnification but might also increase distortion or reduce the field of view.
Knowing the focal length can help in choosing the right lenses for your microscope setup depending on your observations.
Eyepiece Power
The eyepiece power, also known as the diopter, influences how much the eyepiece lens itself magnifies the image created by the objective lens. This power is related to the focal length of the eyepiece lens.
  • Eyepiece power is inversely related to the focal length: \( P = \frac{1}{f_e} \) where \( f_e \) is the eyepiece focal length in meters.
  • To enhance clarity, the eyepiece power often needs to be precisely calculated for the desired total magnification.For example, in the exercise, we calculated \( P \approx -80.8 \mathrm{~D} \), which means the eyepiece contributes a significant amount of magnification itself.
  • A well-chosen eyepiece power can reduce eye strain and improve the detail visible through the microscope.
Choosing the correct eyepiece power ensures that the final image observed is clear, precise, and accurately magnified.

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Most popular questions from this chapter

A terrestrial telescope has three lenses: an objective, an erecting lens, and an eyepiece. (a) Does the erecting lens (1) increase the magnification, (2) increase the physical length of the telescope, (3) decrease the magnification, or (4) decrease the physical length of the telescope? Explain. (b) This terrestrial telescope has focal lengths of \(40 \mathrm{~cm}, 10 \mathrm{~cm},\) and \(5.0 \mathrm{~cm}\) for the objective, erecting lens, and eyepiece, respectively. What is the magnification of the telescope for an object at infinity? (c) What is the length of the telescope barrel?

A modern microscope is equipped with a turret that has three objectives with focal lengths of \(16 \mathrm{~mm}\), \(4.0 \mathrm{~mm},\) and \(1.6 \mathrm{~mm}\) and interchangeable eyepieces of \(5.0 \times\) and \(10 \times .\) A specimen is positioned such that each objective produces an image \(150 \mathrm{~mm}\) from the objective. (a) Which objective and eyepiece combination would you use if you want to have the greatest magnification? How about the least magnification? Explain. (b) What are the greatest and least magnifications possible?

A telescope has an angular magnification of \(-50 \times\) and a barrel \(1.02 \mathrm{~m}\) long. What are the focal lengths of the objective and the eyepiece?

A nearsighted man wears eyeglasses whose lenses have a focal length of \(-0.25 \mathrm{~m}\). How far away is his far point?

\(A-150 \times\) microscope has an eyepiece whose focal length is \(4.4 \mathrm{~cm}\). If the distance between the lenses is \(20 \mathrm{~cm},\) find the focal length of the objective.
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