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Ch 35: Optical Instruments
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 35, Problem 39b

Modern microscopes are more likely to use a camera than human viewing. This is accomplished by replacing the eyepiece in Figure 35.14 with a photo-ocular that focuses the of the objective to a real on the sensor of a digital camera. Suppose the sensor is 22.5 mm wide, a typical value, with 4.0 μm x 4.0 μm pixels. The photo of a cell is 120 pixels in diameter. What is the cell’s actual diameter, in μm?

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1
Determine the size of one pixel in micrometers (μm). The problem states that each pixel is 4.0 μm x 4.0 μm, so the size of one pixel is 4.0 μm along each dimension.
Calculate the total diameter of the cell in micrometers by multiplying the number of pixels (120 pixels) by the size of one pixel (4.0 μm). Use the formula: \( \text{Cell diameter} = \text{Number of pixels} \times \text{Size of one pixel} \).
Substitute the given values into the formula: \( \text{Cell diameter} = 120 \times 4.0 \).
Perform the multiplication to find the cell's actual diameter in micrometers. This step will give you the final result.
Verify the units of the final answer to ensure it is expressed in micrometers (μm), as required by the problem.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Pixel Size and Resolution

Pixel size refers to the dimensions of individual pixels on a sensor, which in this case is 4.0 μm x 4.0 μm. The resolution of an image is determined by the number of pixels it contains; more pixels can capture finer details. Understanding pixel size is crucial for converting pixel measurements into real-world dimensions.
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Magnification

Magnification is the process of enlarging the appearance of an object through optical instruments like microscopes. It is defined as the ratio of the size of the image to the actual size of the object. In this context, knowing the magnification allows us to relate the size of the image (in pixels) to the actual size of the cell.
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Image Measurement Conversion

Image measurement conversion involves translating measurements taken in pixels to actual physical dimensions. This requires knowledge of the sensor's pixel size and the magnification used. By applying these concepts, one can calculate the actual size of an object based on its representation in pixels on the sensor.
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Related Practice
Textbook Question

A microscope with a tube length of 180 mm achieves a total magnification of 800x with a 40x objective and a 20x eyepiece. The microscope is focused for viewing with a relaxed eye. Approximately how far is the sample from the objective lens?

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Textbook Question

A simple and relatively inexpensive microscope eyepiece is the Ramsden eyepiece shown in FIGURE P35.40. Two plano-convex lenses have their curved surfaces facing each other, which a more advanced analysis shows is the orientation that minimizes spherical aberration. That same analysis finds that chromatic aberration is minimized with lens spacing L = 1/2 (f₁ + f₂). Your task is to design a 10x Ramsden eyepiece in which the first lens has a focal length of 30 mm. What are (a) the focal length and (b) the spacing of the second lens?

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Textbook Question

Mars (6800 km diameter) is viewed through a telescope on a night when it is 1.1 x 10⁸ km from the earth. Its angular size as seen through the eyepiece is 0.50°, the same size as the full moon seen by the naked eye. If the eyepiece focal length is 25 mm, how long is the telescope?

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Textbook Question

White light is incident onto a 30° prism at the 40° angle shown in FIGURE P35.41. Violet light emerges perpendicular to the rear face of the prism. The index of refraction of violet light in this glass is 2.0% larger than the index of refraction of red light. At what angle Φ does red light emerge from the rear face?

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Textbook Question

Marooned on a desert island and with a lot of time on your hands, you decide to disassemble your glasses to make a crude telescope with which you can scan the horizon for rescuers. Luckily you’re farsighted, and, like most people, your two eyes have different lens prescriptions. Your left eye uses a lens of power +4.5 D, and your right eye’s lens is +3.0 D. Which lens should you use for the objective and which for the eyepiece? Explain.

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Textbook Question

High-power lasers are used to cut and weld materials by focusing the laser beam to a very small spot. This is like using a magnifying lens to focus the sun's light to a small spot that can burn things. As an engineer, you have designed a laser cutting device in which the material to be cut is placed 5.0 cm behind the lens. You have selected a high-power laser with a wavelength of 1.06 μm. Your calculations indicate that the laser must be focused to a 5.0-μm-diameter spot in order to have sufficient power to make the cut. What is the minimum diameter of the lens you must install?

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