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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 pinhole camera is just a rectangular box with a tiny hole in one face. The film is on the face opposite this hole, and that is where the image is formed. The camera forms an image without a lens. (a) Make a clear ray diagram to show how a pinhole camera can form an image on the film without using a lens. (Hint: Put an object outside the hole, and then draw rays passing through the hole to the opposite side of the box.) (b) A certain pinhole camera is a box that is \(25 \mathrm{~cm}\) square and \(20.0 \mathrm{~cm}\) deep, with the hole in the middle of one of the \(25 \mathrm{~cm} \times 25 \mathrm{~cm}\) faces. If this camera is used to photograph a fierce chicken that is \(18 \mathrm{~cm}\) high and \(1.5 \mathrm{~m}\) in front of the camera, how large is the image of this bird on the film? What is the lateral magnification of this camera?

A coin is placed next to the convex side of a thin spherical glass shell having a radius of curvature of \(18.0 \mathrm{~cm} .\) Reflection from the surface of the shell forms an image of the \(1.5-\mathrm{cm}\) -tall coin that is \(6.00 \mathrm{~cm}\) behind the glass shell. Where is the coin located? Determine the size, orientation, and nature (real or virtual) of the image.

The smallest object we can resolve with our eye is limited by the size of the light receptor cells in the retina. In order for us to distinguish any detail in an object, its image cannot be any smaller than a single retinal cell. Although the size depends on the type of cell (rod or cone), a diameter of a few microns \((\mu \mathrm{m})\) is typical near the center of the eye. We shall model the eye as a sphere \(2.50 \mathrm{~cm}\) in diameter with a single thin lens at the front and the retina at the rear, with light receptor cells \(5.0 \mu \mathrm{m}\) in diameter. (a) What is the smallest object you can resolve at a near point of \(25 \mathrm{~cm} ?\) (b) What angle is subtended by this object at the eye? Express your answer in units of minutes \(\left(1^{\circ}=60 \mathrm{~min}\right),\) and compare it with the typical experimental value of about \(1.0 \mathrm{~min} .\) (Note: There are other limitations, such as the bending of light as it passes through the pupil, but we shall ignore them here.)

Two thin lenses with a focal length of magnitude \(12.0 \mathrm{~cm},\) the first diverging and the second converging, are located \(9.00 \mathrm{~cm}\) apart. An object \(2.50 \mathrm{~mm}\) tall is placed \(20.0 \mathrm{~cm}\) to the left of the first (diverging) lens. (a) How far from this first lens is the final image formed? (b) Is the final image real or virtual? (c) What is the height of the final image? Is it erect or inverted? (Hint: See the preceding two exercises.)

It is your first day at work as a summer intern at an optics company. Your supervisor hands you a diverging lens and asks you to measure its focal length. You know that with a converging lens, you can measure the focal length by placing an object a distance \(s\) to the left of the lens, far enough from the lens for the image to be real, and viewing the image on a screen that is to the right of the lens. By adjusting the position of the screen until the image is in sharp focus, you can determine the image distance \(s^{\prime}\) and then use Eq. (34.16) to calculate the focal length \(f\) of the lens. But this procedure won't work with a diverging lens - by itself, a diverging lens produces only virtual images, which can't be projected onto a screen. Therefore, to determine the focal length of a diverging lens, you do the following: First you take a converging lens and measure that, for an object \(20.0 \mathrm{~cm}\) to the left of the lens, the image is \(29.7 \mathrm{~cm}\) to the right of the lens. You then place a diverging lens \(20.0 \mathrm{~cm}\) to the right of the converging lens and measure the final image to be \(42.8 \mathrm{~cm}\) to the right of the converging lens. Suspecting some inaccuracy in measurement, you repeat the lenscombination measurement with the same object distance for the converging lens but with the diverging lens \(25.0 \mathrm{~cm}\) to the right of the converging lens. You measure the final image to be \(31.6 \mathrm{~cm}\) to the right of the converging lens. (a) Use both lens-combination measurements to calculate the focal length of the diverging lens. Take as your best experimental value for the focal length the average of the two values. (b) Which position of the diverging lens, \(20.0 \mathrm{~cm}\) to the right or \(25.0 \mathrm{~cm}\) to the right of the converging lens, gives the tallest image?

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