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Chapter 31: Problem 68

A 300 -power compound microscope has a 4.5 -mm-focal-length objective lens. If the distance from objective to eyepiece is \(10 \mathrm{cm},\) what should be the focal length of the eyepiece?

Short Answer

Expert verified
The focal length of the eyepiece should be 23.5 mm.

Step by step solution

01

Determine the equation for total magnification

The total magnification of a compound microscope is given by the formula: \(M = 1 + D/f_{\mathrm{objective}})\) * \(250/f_{\mathrm{eyepiece}})\),where M is the total magnification, D = 25 cm (the near-point distance), \(f_{\mathrm{objective}}\) is the focal length of the objective lens, and \(f_{\mathrm{eyepiece}}\) is the focal length of the eyepiece. Here the challenge is to find \(f_{\mathrm{eyepiece}}\) when M, D, and \(f_{\mathrm{objective}}\) are given.
02

Substitute the given values into the equation

Substitute \( M = 300\), \(D = 25 \, \mathrm{cm} = 0.25 \, \mathrm{m}\), and \(f_{\mathrm{objective}} = 4.5 \, \mathrm{mm} = 0.0045 \, \mathrm{m}\) into the equation:\(300 = 1 + \frac{0.25}{0.0045}\) * \(\frac{250}{f_{\mathrm{eyepiece}}}\)
03

Solve for eyepiece focal length

First, simplify the equation:\[299 = \frac{0.25}{0.0045} * \frac{250}{f_{\mathrm{eyepiece}}}\]Then, isolate \(f_{\mathrm{eyepiece}}\) on one side of the equation:\[f_{\mathrm{eyepiece}} = \frac{0.25}{0.0045} * \frac{250}{299}\]When you perform the multiplication and division, you'll find that\[f_{\mathrm{eyepiece}} = 0.0235 \, \mathrm{m}\] or \[f_{\mathrm{eyepiece}} = 23.5 \, \mathrm{mm}\]

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Most popular questions from this chapter

A contact lens prescription calls for +2.25 -diopter lenses with inner curvature radius \(8.6 \mathrm{mm}\) to fit the patient's cornea. (a) If the lenses are plastic with \(n=1.56,\) what should be the outer curvature radius? (b) With these lenses, the patient comfortably reads a newspaper \(30 \mathrm{cm}\) from her eyes. Where's the image as viewed through the lenses?

Under what circumstances will the image in a concave mirror be the same size as the object?

The maximum magnification of a simple magnifier occurs with the image at the 25 -cm near point. Show that the angular magnification is \(m=1+(25 \mathrm{cm} / \mathrm{f}),\) where \(f\) is the focal length.

A candle is on the axis of a 15 -cm-focal-length concave mirror, \(36 \mathrm{cm}\) from the mirror. (a) Where is its image? (b) How do the image and object sizes compare? (c) Is the image real or virtual?

An object and its lens-produced real image are \(2.4 \mathrm{m}\) apart. If the lens has 55 -cm focal length, what are the possible values for the object distance and magnification?
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