/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 65 The overall angular magnificatio... [FREE SOLUTION] | 91Ó°ÊÓ

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The overall angular magnification of a microscope is \(M=-178 .\) The eyepiece has focal length \(15.0 \mathrm{~mm}\) and the final image is at infinity. The separation between the two lenses is \(202 \mathrm{~mm}\). What is the focal length of the objective? Do not use the approximation \(s_{1} \approx f_{1}\) in the expression for \(M\).

Short Answer

Expert verified
The focal length of the objective lens is approximately 0.7 mm.

Step by step solution

01

Derive the magnification formula for microscope

The magnification for a simple microscope is given by \(M=-\frac{L-f_2}{f_1f_2}\), where \(f_1\) is the focal length of the objective, \(f_2\) is the focal length of the eyepiece, and \(L\) is the distance between the object and image. Here, negative sign indicates that the final image is inverted.
02

Substitute given values into magnification formula

Substitute \(M=-178\), \(f_2=15mm\) and \(L=202mm\) into the formula to be rewritten as: \(-178=-\frac{202-15}{f_1*15}\) or \(178=\frac{187}{f_1*15}\).
03

Solve for the focal length of the objective lens

Rearranging for \(f_1\) and converting everything to the same units (mm), we get \(f_1=\frac{187}{178*15}\) in mm. Calculating the above expression gives the value of \(f_1\) to be approximately \(0.7mm\).

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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 measure of how much larger an object appears when viewed through an optical instrument compared to when viewed with the naked eye. For a microscope, it is determined by the combination of the objective lens and the eyepiece. This magnification accounts for the apparent increase in the size of the object. It is especially important in applications such as biology and materials science where the details of tiny structures need to be examined closely. When calculating angular magnification in microscopes, you often encounter a negative sign. This indicates that the final image appears inverted. The formula for angular magnification of a microscope is given by:\[ M = - \frac{L - f_2}{f_1 f_2} \]where:- \(M\) is the angular magnification,- \(L\) is the separation between the lenses,- \(f_1\) is the focal length of the objective lens,- \(f_2\) is the focal length of the eyepiece lens. In the solution provided, the negative angular magnification of \(-178\) highlights the considerable magnification power of the optical setup, although the image produced is inverted.
Focal Length
Focal length is a critical parameter in lens design, defining the distance over which parallel rays of light are brought to a focus. In microscopes, each lens, objective and eyepiece, has a specific focal length that determines how it bends light and contributes to the overall magnification. The focal length of the eyepiece in our example is given as 15 mm. This value, typical for eyepieces, indicates that it focuses light at a relatively short distance, playing a role in how the image appears via the eyepiece. The focal length of the objective lens, calculated to be approximately 0.7 mm in this exercise, is even shorter. This short focal length is crucial because it allows for significant magnification at the first stage of the microscope's optics. The combined effect of these two focal lengths in the magnification formula affects how the lenses enlarge an object, making details visible that would otherwise be missed with the naked eye.
Objective Lens
The objective lens is one of the key components of a microscope. It is located closest to the specimen that is being observed and is responsible for the primary magnification of the image. This lens gathers light from the specimen and magnifies the image considerably. In the given exercise, the goal was to determine the focal length of the objective lens from the known values of the microscope's total angular magnification and the eyepiece focal length. Calculating a focal length of approximately 0.7 mm, we see that the objective lens in this setup is designed for high magnification with a very short focal length. Key characteristics of objective lenses:
  • Short focal length: Increases the magnification power.
  • Proximity to the object: Needs to be close to the specimen for effective magnification.
  • High numerical aperture: Allows more light to be collected for a brighter image.
Eyepiece
The eyepiece, also known as the ocular lens, is the lens you look through in a microscope. It provides the final stage of magnification before the observer views the image. Located further from the specimen, it works in tandem with the objective lens to produce an enlarged image of the specimen. In this particular problem, the eyepiece has a focal length of 15 mm. The role of the eyepiece is to magnify the image produced by the objective lens and present it to the observer at a comfortable viewing angle. This final image is often projected to infinity, making it easier for the human eye to view without straining. The eyepiece does not just magnify the image; it also contributes to the apparent field of view and the comfort of viewing, which are important for prolonged observation through the microscope. The combination of focal lengths of the eyepiece and the objective lens directly influences the overall magnification of the microscope.
Image Formation
Image formation in a microscope occurs through a two-stage process involving both the objective lens and the eyepiece. Light from the specimen passes through the objective lens, which generates a magnified image of the specimen. This primary image is real, inverted, and is the first major magnification stage. The eyepiece then takes this initial image and magnifies it even more to produce a secondary virtual image that is viewed by the user. This is a key aspect of microscopes, as understanding how images are formed helps in setting up the device correctly and comprehending the orientation of the observed images:
  • Initial image by the objective: Real and inverted.
  • Final image by the eyepiece: Virtual and can be projected at infinity.
This process allows significant magnification and is why microscopes can reveal details that are invisible to the naked eye.

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Most popular questions from this chapter

A converging lens with a focal length of \(9.00 \mathrm{~cm}\) forms an image of a \(4.00-\mathrm{mm}\) -tall real object that is to the left of the lens. The image is \(1.30 \mathrm{~cm}\) tall and erect. Where are the object and image located? Is the image real or virtual?

You wish to project the image of a slide on a screen \(9.00 \mathrm{~m}\) from the lens of a slide projector. (a) If the slide is placed \(15.0 \mathrm{~cm}\) from the lens, what focal length lens is required? (b) If the dimensions of the picture on a \(35 \mathrm{~mm}\) color slide are \(24 \mathrm{~mm} \times 36 \mathrm{~mm},\) what is the minimum size of the projector screen required to accommodate the image?

When an object is placed at the proper distance to the left of a converging lens, the image is focused on a screen \(30.0 \mathrm{~cm}\) to the right of the lens. A diverging lens is now placed \(15.0 \mathrm{~cm}\) to the right of the converging lens, and it is found that the screen must be moved \(19.2 \mathrm{~cm}\) farther to the right to obtain a sharp image. What is the focal length of the diverging lens?

A camera with a 90-mm-focal-length lens is focused on an object \(1.30 \mathrm{~m}\) from the lens. To refocus on an object \(6.50 \mathrm{~m}\) from the lens, by how much must the distance between the lens and the sensor be changed? To refocus on the more distant object, is the lens moved toward or away from the sensor?

The cornea of the eye has a radius of curvature of approximately \(0.50 \mathrm{~cm},\) and the aqueous humor behind it has an index of refraction of \(1.35 .\) The thickness of the cornea itself is small enough that we shall neglect it. The depth of a typical human eye is around \(25 \mathrm{~mm}\). (a) What would have to be the radius of curvature of the cornea so that it alone would focus the image of a distant mountain on the retina, which is at the back of the eye opposite the cornea? (b) If the cornea focused the mountain correctly on the retina as described in part (a), would it also focus the text from a computer screen on the retina if that screen were \(25 \mathrm{~cm}\) in front of the eye? If not, where would it focus that text: in front of or behind the retina? (c) Given that the cornea has a radius of curvature of about \(5.0 \mathrm{~mm}\), where does it actually focus the mountain? Is this in front of or behind the retina? Does this help you see why the eye needs help from a lens to complete the task of focusing?

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