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Chapter 24: Problem 40

A microscope has an eyepicce that gives an angular magnification of 5.00 for a final image at infinity and an objective lens of focal length $15.0 \mathrm{mm}\(. The tube length of the microscope is \)16.0 \mathrm{cm} .$ (a) What is the transverse magnification due to the objective lens alone? (b) What is the angular magnification due to the microscope? (c) How far from the objective should the object be placed?

Short Answer

Expert verified
Answer: (a) The transverse magnification of the objective lens is 8.665. (b) The angular magnification of the whole microscope is 43.325. (c) The object should be placed at a distance of 1.673 cm from the objective lens.

Step by step solution

01

Calculate the image distance for the objective lens

Given the tube length of the microscope as \(16.0 \mathrm{cm}\) and the focal length of the objective lens as \(15.0 \mathrm{mm}\), we will use the lens formula to find the image distance \((v)\) for the objective lens. The lens formula is given by: \(\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\) First, we need to convert the focal length to centimeters: \(15.0 \mathrm{mm}= 1.5 \mathrm{cm}\). Now we can solve the lens formula for the image distance \((v)\). However, for a final image at infinity, we know that the image formed by the objective should be at the focal point of the eyepiece. Since the tube length \(L = 16.0 \mathrm{cm}\), the image distance \(v = 16.0 - 1.5 = 14.5 \mathrm{cm}\).
02

Calculate the transverse magnification of the objective lens

To find the transverse magnification of the objective lens (\(M_{obj}\)), we'll use the formula: \(M_{obj} = \frac{v}{u}\) We already know the image distance \(v = 14.5 \mathrm{cm}\). Now we need to find the object distance \((u)\). Rearranging the lens formula, we get: \(u = \frac{1}{\frac{1}{f} - \frac{1}{v}}\) Substituting the values of \(f\) and \(v\), we get: \(u = \frac{1}{\frac{1}{1.5} - \frac{1}{14.5}} = \frac{1}{\frac{13}{21.75}} = 1.673 \mathrm{cm}\) Now we can calculate the transverse magnification of the objective lens: \(M_{obj} = \frac{14.5}{1.673} = 8.665\) So, the transverse magnification due to the objective lens alone is \(8.665\).
03

Calculate the angular magnification of the whole microscope

Since the angular magnification of the eyepiece (\(M_{eye}\)) is given as \(5.00\), we can calculate the angular magnification of the entire microscope (\(M_{total}\)) as the product of the magnifications due to the objective lens and the eyepiece: \(M_{total} = M_{obj} \cdot M_{eye} = 8.665 \cdot 5.00 = 43.325\) The angular magnification due to the microscope is \(43.325\).
04

Find the distance between the object and the objective lens

The object distance \((u)\) is already calculated in Step 2 as \(1.673 \mathrm{cm}\). This is the distance at which the object should be placed from the objective lens. In summary: (a) The transverse magnification due to the objective lens alone is \(8.665\). (b) The angular magnification due to the microscope is \(43.325\). (c) The object should be placed at a distance of \(1.673 \mathrm{cm}\) from the objective lens.

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Most popular questions from this chapter

A dissecting microscope is designed to have a large distance between the object and the objective lens. Suppose the focal length of the objective of a dissecting microscope is \(5.0 \mathrm{cm},\) the focal length of the eyepiece is \(4.0 \mathrm{cm},\) and the distance between the lenses is $32.0 \mathrm{cm} .$ (a) What is the distance between the object and the objective lens? (b) What is the angular magnification?
An object is placed \(12.0 \mathrm{cm}\) in front of a lens of focal length $5.0 \mathrm{cm} .\( Another lens of focal length \)4.0 \mathrm{cm}\( is placed \)2.0 \mathrm{cm}$ past the first lens. (a) Where is the final image? Is it real or virtual? (b) What is the overall magnification? (interactive: virtual optics lab).
A photographer wishes to take a photograph of the Eiffel Tower $(300 \mathrm{m}\( tall ) from across the Seine River, a distance of \)300 \mathrm{m}$ from the tower. What focal length lens should she use to get an image that is 20 mm high on the film?
Esperanza uses a 35 -mm camera with a standard lens of focal length $50.0 \mathrm{mm}\( to take a photo of her son Carlos, who is \)1.2 \mathrm{m}$ tall and standing \(3.0 \mathrm{m}\) away. (a) What must be the distance between the lens and the film to get a properly focused picture? (b) What is the magnification of the image? (c) What is the height of the image of Carlos on the film?
The eyepiece of a microscope has a focal length of \(1.25 \mathrm{cm}\) and the objective lens focal length is \(1.44 \mathrm{cm} .\) (a) If the tube length is \(18.0 \mathrm{cm},\) what is the angular magnification of the microscope? (b) What objective focal length would be required to double this magnification?
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