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Chapter 24: Problem 44

Use the thin-lens equation to show that the transverse magnification due to the objective of a microscope is \(m_{o}=-U f_{o} .\) [Hints: The object is near the focal point of the objective: do not assume it is at the focal point. Eliminate \(p_{0}\) to find the magnification in terms of \(q_{0}\) and $\left.f_{0} . \text { How is } L \text { related to } q_{0} \text { and } f_{0} ?\right].$

Short Answer

Expert verified
Answer: The transverse magnification due to the objective of a microscope is given by \(m_o = -Uf_o\), where \(U = \frac{L-f_e}{f_o}\), \(L\) is the tube length, and \(f_o\) and \(f_e\) are the focal lengths of the objective and eyepiece, respectively.

Step by step solution

01

Define the thin-lens equation

The thin-lens equation is given by \(\frac{1}{f} = \frac{1}{p} + \frac{1}{q}\), where \(f\) is the focal length, \(p\) is the object distance, and \(q\) is the image distance.
02

Apply the lens equation to the microscope objective

For the microscope objective, we can write the lens equation as \(\frac{1}{f_o} = \frac{1}{p_o} + \frac{1}{q_o}\), where \(f_o\) is the objective focal length, \(p_o\) is the object distance, and \(q_o\) is the image distance.
03

Rearrange the equation to eliminate \(p_o\)

To eliminate \(p_o\), we can rearrange the equation as \(p_o = \frac{f_o q_o}{q_o - f_o}\).
04

Calculate the transverse magnification of the objective

The transverse magnification formula is \(m_o = \frac{-q_o}{p_o}\). Using the rearranged equation from Step 3, we get \(m_o = \frac{-q_o}{\frac{f_o q_o}{q_o - f_o}} = - \frac{q_o (q_o - f_o)}{f_o q_o}= -\frac{q_o - f_o}{f_o}\).
05

Find the relationship between \(L\), \(q_o\), and \(f_o\)

To find the relationship between \(L\), \(q_o\), and \(f_o\), we need to consider the geometry of the microscope. The tube length \(L\) is the distance between the objective and eyepiece focal points. As the object is near the focal point of the objective, we have \(L \approx q_o + f_e - f_o\). We can rearrange this expression for \(q_o\) as \(q_o = L + f_o - f_e\), where \(f_e\) is the focal length of the eyepiece.
06

Substitute the relationship into the magnification formula

Substituting the relationship from Step 5 into the magnification formula in Step 4, we have \(m_o = -\frac{(L + f_o - f_e) - f_o}{f_o} = -\frac{L-f_e}{f_o}\).
07

Show the final expression for the transverse magnification

To express the transverse magnification due to the objective of a microscope, we can now write \(m_o = -Uf_o\), where \(U = \frac{L-f_e}{f_o}\).

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Most popular questions from this chapter

You have two lenses of focal length \(25.0 \mathrm{cm}\) (lens 1 ) and $5.0 \mathrm{cm}$ (lens 2 ). (a) To build an astronomical telescope that gives an angular magnification of \(5.0,\) how should you use the lenses (which for objective and which for eyepiece)? Explain. (b) How far apart should they be?
A slide projector has a lens of focal length \(12 \mathrm{cm} .\) Each slide is \(24 \mathrm{mm}\) by \(36 \mathrm{mm}\) (see the figure with Problem 16). The projector is used in a room where the screen is \(5.0 \mathrm{m}\) from the projector. How large must the screen be?
An object is located \(10.0 \mathrm{cm}\) in front of a converging lens with focal length \(12.0 \mathrm{cm} .\) To the right of the converging lens is a second converging lens, \(30.0 \mathrm{cm}\) from the first lens, of focal length \(10.0 \mathrm{cm} .\) Find the location of the final image by ray tracing and verify by using the lens equations.
A microscope has an eyepiece of focal length \(2.00 \mathrm{cm}\) and an objective of focal length \(3.00 \mathrm{cm} .\) The eyepiece produces a virtual image at the viewer's near point \((25.0 \mathrm{cm}\) from the eye). (a) How far from the eyepiece is the image formed by the objective? (b) If the lenses are \(20.0 \mathrm{cm}\) apart, what is the distance from the objective lens to the object being viewed? (c) What is the angular magnification?
A refracting telescope has an objective lens with a focal length of $2.20 \mathrm{m}\( and an eyepiece with a focal length of \)1.5 \mathrm{cm} .$ If you look through this telescope the wrong way, that is, with your eye placed at the objective lens, by what factor is the angular size of an observed object reduced?
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