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A certain microscope is provided with objectives that have focal lengths of \(16 \mathrm{~mm}, 4 \mathrm{~mm},\) and \(1.9 \mathrm{~mm}\) and with eyepieces that have angular magnifications of \(5 \times\) and \(10 \times .\) Each objective forms an image \(120 \mathrm{~mm}\) beyond its second focal point. Determine (a) the largest overall angular magnification obtainable and (b) the smallest overall angular magnification obtainable.

Short Answer

Expert verified
The largest angular magnification is approximately 631.6, and the smallest is 37.5.

Step by step solution

01

Understanding Angular Magnification

The angular magnification of a microscope is determined by the product of the magnification of the objective lens and the magnification of the eyepiece lens. Hence, we have \( M = M_{objective} \times M_{eyepiece} \), where \( M_{objective} \) is the magnification of the objective lens, and \( M_{eyepiece} \) is the angular magnification of the eyepiece.
02

Calculating Magnification of Objectives

For an objective lens, the magnification is given by \( M_{objective} = \frac{L}{f_{objective}} \), where \( L = 120 \mathrm{~mm} \) is the image distance and \( f_{objective} \) is the focal length of the objective lens. We calculate for each lens:- For \(16 \mathrm{~mm}\) lens: \( M_{objective} = \frac{120}{16} = 7.5 \)- For \(4 \mathrm{~mm}\) lens: \( M_{objective} = \frac{120}{4} = 30 \)- For \(1.9 \mathrm{~mm}\) lens: \( M_{objective} = \frac{120}{1.9} \approx 63.16 \)
03

Identifying Eyepiece Magnifications

The eyepieces have angular magnifications given in the problem: \( M_{eyepiece1} = 5 \times \) and \( M_{eyepiece2} = 10 \times \). These are used directly in combination with the respective objective lens magnifications to find the overall magnifications.
04

Calculating Largest Angular Magnification

To find the largest overall angular magnification, pair the largest objective magnification with the largest eyepiece magnification: \[ M_{max} = 63.16 \times 10 = 631.6 \] Thus, the largest angular magnification obtainable is approximately 631.6.
05

Calculating Smallest Angular Magnification

To find the smallest overall angular magnification, pair the smallest objective magnification with the smallest eyepiece magnification:\[ M_{min} = 7.5 \times 5 = 37.5 \] Thus, the smallest angular magnification obtainable is 37.5.

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Key Concepts

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

Angular Magnification
Angular magnification is a key concept in understanding how microscopes function, as it pertains to how much larger an image appears compared to the actual object. This is determined by the combination of the magnifications provided by the microscope's objective and eyepiece lenses. By multiplying these magnifications, we obtain the angular magnification, given by the formula:
  • \( M = M_{objective} \times M_{eyepiece} \)
Essentially, this product reflects how effectively the microscope can enlarge images, making even the smallest details visible to the observer.
Objective Lens Magnification
Objective lens magnification is a crucial factor in the overall performance of a microscope. This lens is responsible for the initial enlargement of the specimen. The magnification provided by the objective lens is calculated using the formula:
  • \( M_{objective} = \frac{L}{f_{objective}} \)
where \( L \) is the image distance, or the distance from the second focal point to the image, and \( f_{objective} \) is the focal length of the objective lens. Longer \( L \) or shorter \( f_{objective} \) values result in higher magnifications. In our example, the objective lens magnifications were computed for lenses with focal lengths of 16 mm, 4 mm, and 1.9 mm.
Eyepiece Magnification
The eyepiece or ocular lens additionally magnifies the image produced by the objective lens. Eyepieces are often rated by their angular magnification values, which indicate how much they can further enlarge the already magnified image.
  • For instance, if an eyepiece has a magnification of \(5 \times\), it further enlarges the image size fivefold.
In our example, the eyepieces provided have magnifications of \(5 \times\) and \(10 \times\), directly affecting the overall angular magnification when combined with the objective lens magnification.
Focal Length Calculation
Understanding focal length is essential when working with any optical instruments like microscopes. The focal length of a lens, denoted as \( f \), is the distance from the lens to the point where light rays converge to form a sharp image. In the formula for objective magnification, \( M_{objective} = \frac{L}{f_{objective}} \), we see that the focal length directly influences how much the lens can magnify the object.
  • A shorter focal length results in a higher magnification, as seen with the 1.9 mm focal length objective yielding a greater magnification.
This relationship highlights why understanding and calculating focal length is vital.
Image Distance
Image distance, denoted as \( L \), is the distance between the lens and the location where the image is focused. It's a crucial element in calculating the magnifications achieved by the objective lens in a microscope. The formula \( M_{objective} = \frac{L}{f_{objective}} \) incorporates this image distance.
  • If \( L \) increases, the total magnification from the objective lens increases too.
For optimal microscopes, the image distance is fixed to allow for precise calculations and effective magnification that produces clear images.
Microscope Optics
Microscope optics refers to the integrated system of lenses that magnifies and resolves fine details of specimens. By standard design, a microscope consists of an objective lens and an eyepiece.
  • The objective lens captures and enlarges an initial image of the specimen.
  • The eyepiece then magnifies this image further for detailed examination.
In terms of design, it's crucial to balance the focal lengths and placements, ensuring optimal image quality and suitable magnification power for practical use. This combination and the interplay of objective and eyepiece lenses are what create the detailed images that make microscopes indispensable in fields such as biology and material science.

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Most popular questions from this chapter

An amateur photographer purchases a vintage camera at a flea market. In order to determine the focal length of the camera's lens, he sets a soup can at various distances from the camera. At each distance he brings the camera into focus on the can and then carefully measures the distance between the lens and the film. His measurements are given in the table. $$ \begin{array}{cc} \hline \text { Can distance }(\mathrm{cm}) & \text { Lens distance }(\mathrm{mm}) \\ \hline 25 & 118 \\ 35 & 104 \\ 50 & 95 \\ 75 & 89 \\ 100 & 87 \\ \hline \end{array} $$ Make a plot of the inverse lens distance as a function of the inverse can distance. Using a linear "best fit" to the data, determine the focal length of the lens.

Your digital camera has a lens with a \(50 \mathrm{~mm}\) focal length and a sensor array that measures \(4.82 \mathrm{~mm} \times 3.64 \mathrm{~mm}\). Suppose you're at the zoo and want to take a picture of a 4.50 -m-tall giraffe. If you want the giraffe to exactly fit the longer dimension of your sensor array, how far away from the animal will you have to stand?

Contact lenses are placed right on the eyeball, so the distance from the eye to an object (or image) is the same as the distance from the lens to that object (or image). A certain person can see distant objects well, but his near point is \(45.0 \mathrm{~cm}\) from his eyes instead of the usual \(25.0 \mathrm{~cm}\). (a) Is this person nearsighted or farsighted? (b) What type of lens (converging or diverging) is needed to correct his vision? (c) If the correcting lenses will be contact lenses, what focal-length lens is needed and what is its power in diopters?

A telescope is constructed from two lenses with focal lengths of \(95.0 \mathrm{~cm}\) and \(15.0 \mathrm{~cm},\) the \(95.0-\mathrm{cm}\) lens being used as the objective. Both the object being viewed and the final image are at infinity. (a) Find the angular magnification of the telescope. (b) Find the height of the image formed by the objective of a building 60.0 \(\mathrm{m}\) tall and \(3.00 \mathrm{~km}\) away. \((\mathrm{c}) \mathrm{What}\) is the angular size of the final image as viewed by an eye very close to the eyepiece?

Zoom lens. A zoom lens is a lens that varies in focal length. The zoom lens on a certain digital camera varies in focal length from \(6.50 \mathrm{~mm}\) to \(19.5 \mathrm{~mm}\). This camera is focused on an object \(2.00 \mathrm{~m}\) tall that is \(1.50 \mathrm{~m}\) from the camera. Find the distance between the lens and the photo sensors and the height of the image (a) when the zoom is set to \(6.50 \mathrm{~mm}\) focal length and (b) when it is at \(19.5 \mathrm{~mm}\). (c) Which is the telephoto focal length, \(6.50 \mathrm{~mm}\) or \(19.5 \mathrm{~mm} ?\)

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