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Chapter 39: Problem 27

In a compound microscope, the focal lengths of the objective and eyepiece are \(+0.50 \mathrm{~cm}\) and \(+2.0 \mathrm{~cm}\), respectively. The instrument is focused on an object \(0.52 \mathrm{~cm}\) from the objective lens. Compute the magnifying power of the microscope if the virtual image is viewed by the eye at a distance of \(25 \mathrm{~cm}\).

Short Answer

Expert verified
The magnifying power of the microscope is approximately 250x.

Step by step solution

01

Calculate the image distance for the objective lens

First, use the lens formula for the objective lens, which is \( \frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o} \), where \( f_o = 0.50 \mathrm{~cm} \) and \( u_o = 0.52 \mathrm{~cm} \). Solve for \( v_o \) which is the image distance formed by the objective:\[ \frac{1}{v_o} = \frac{1}{0.50} + \frac{1}{0.52} \]Calculate \( v_o \).
02

Determine the magnification by the objective

The magnification produced by the objective lens is given by the formula \( M_o = \frac{v_o}{u_o} \). Use the value of \( v_o \) obtained from Step 1 to calculate \( M_o \).
03

Calculate the image distance for the eyepiece lens

Since the image formed by the objective acts as the object for the eyepiece and a virtual image is produced at \( -25 \mathrm{~cm} \), use the lens formula for the eyepiece:\[ \frac{1}{f_e} = \frac{1}{v_e} - \frac{1}{u_e} \]Here, \( v_e = -25 \mathrm{~cm} \) because the image is virtual and \( f_e = 2.0 \mathrm{~cm} \). Solve for \( u_e \).
04

Determine the magnification by the eyepiece

The eyepiece magnification \( M_e \) is given by the formula \( M_e = 1 + \frac{D}{f_e} \), where \( D = 25 \mathrm{~cm} \) is the near point distance. Substitute the given values to find \( M_e \).
05

Calculate the total magnifying power of the microscope

The total magnifying power of the compound microscope is the product of the magnifications produced by the objective and the eyepiece: \( M = M_o \times M_e \). Use the values obtained from Steps 2 and 4 to find \( M \).

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Key Concepts

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

Focal Length
The focal length of a lens is a critical component of its function within a compound microscope. In simple terms, the focal length is the distance between the lens and its focus point, where parallel rays of light converge to form a sharp image.
For a compound microscope, each lens鈥攖he objective and the eyepiece鈥攈as its own unique focal length:
  • Objective Lens: In our example, the focal length is given as +0.50 cm. This short focal length helps capture a detailed and large image of the microscopic specimen.
  • Eyepiece Lens: The eyepiece has a longer focal length of +2.0 cm. This lens enlarges the image formed by the objective lens, aiding in a more comfortable viewing by the observer.
Understanding the focal length of both lenses helps determine how they manipulate light to magnify objects effectively, forming an essential part of the microscope's optics.
This distance influences the clarity and size of the image produced, affecting everything from object detail to user comfort.
Magnification
In the world of compound microscopes, magnification is a pivotal aspect as it indicates how much larger an object appears compared to its actual size.
Magnification is achieved by the compound action of two lenses:
  • Objective Lens Magnification: The objective lens's ability to zoom in on an object is represented by \( M_o = \frac{v_o}{u_o} \), where \( v_o \) is the image distance from the lens and \( u_o \) is the object distance.
  • Eyepiece Lens Magnification: The eyepiece provides additional amplification with the formula \( M_e = 1 + \frac{D}{f_e} \). Here, \( D = 25 \mathrm{~cm}\) is the standard distance for clear vision.
  • Total Magnification: The total is calculated by combining both, \( M = M_o \times M_e \), to show how powerful the microscope's magnifying ability is.
This multiply-stacked system allows tiny details in small objects to be seen clearly thus making it a powerful tool in scientific observation.
Lens Formula
The lens formula is a universal equation used to determine the relationship between the object distance (\( u \)), the image distance (\( v \)), and the focal length (\( f \)) of a lens. It is represented by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Each part of this equation holds significance in calculating how lenses focus light:
  • Objective Lens: With known values of object distance and focal length, this formula helps compute the position of the intermediate image produced by the objective lens.
  • Eyepiece Lens: Knowing the eyepiece's focal length and desired virtual image location assists in determining the object's placement relative to this lens.
This lens formula is integral in ensuring that the images are sharply focused and correctly positioned for comfortable viewing using a compound microscope.
Virtual Image
In optics, a virtual image is an image that appears to be located behind the lens and cannot be projected onto a screen. In compound microscopes, the eyepiece lens is crucial for forming virtual images.
When you look into a microscope:
  • The objective lens first forms a real image by focusing light onto a specific point. This image acts as the object for the eyepiece.
  • The eyepiece lens then magnifies this real image into a larger virtual image.
  • Because the virtual image is situated at a distance that is easy for the human eye to focus on, observing such images becomes more pleasant and less straining.
Virtual images are significant as they allow scientists and students to observe small details with ease, making these microscopes invaluable in various fields of study and research.

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Most popular questions from this chapter

Two positive lenses, having focal lengths of \(+2.0 \mathrm{~cm}\) and \(+5.0 \mathrm{~cm}\), are \(14 \mathrm{~cm}\) apart as shown. An object \(A B\) is placed \(3.0\) \(\mathrm{cm}\) in front of the \(+2.0\) lens. Determine the position and magnification of the final image A"B" formed by this combination of lenses. To locate image \(A^{\prime} B^{\prime}\) formed by the \(+2.0\) lens alone: $$ \frac{1}{s_{i}}=\frac{1}{f}-\frac{1}{s_{o}}=\frac{1}{2.0}-\frac{1}{3.0}=\frac{1}{6.0} \quad \text { or } \quad s_{i}=6.0 \mathrm{~cm} $$ The image \(A^{\prime} B^{\prime}\) is real, inverted, and \(6.0 \mathrm{~cm}\) beyond the \(+2.0\) lens. To locate the final image \(A^{\prime \prime} B^{\prime \prime}\) : The image \(A^{\prime} B^{\prime}\) is \((14-6.0) \mathrm{cm}=\) \(8.0 \mathrm{~cm}\) in front of the \(+5.0\) lens and is taken as a real object for the \(+5.0\) lens. $$ \frac{1}{s_{i}}=\frac{1}{5.0}-\frac{1}{8.0} \quad \text { or } \quad s_{i}=13.3 \mathrm{~cm} $$ \(A^{\prime \prime} B^{\prime \prime}\) is real, erect, and \(13 \mathrm{~cm}\) from the \(+5\) lens. Then, $$ M_{T}=\frac{\overline{A^{\prime \prime} B^{\prime \prime}}}{\overline{A B}}=\frac{\overline{A^{\prime} B^{\prime}}}{\overline{A B}} \times \frac{\overline{A^{\prime \prime} B^{\prime \prime}}}{\overline{A^{\prime} B^{\prime}}}=\frac{6.0}{3.0} \times \frac{13.3}{8.0}=3.3 $$ Note that the magnification produced by a combination of lenses is the product of the individual magnifications.

A microscope has two interchangeable objective lenses (3.0 mm and \(7.0 \mathrm{~mm}\) ) and two interchangeable eyepieces \((3.0 \mathrm{~cm}\) and \(5.0\) \(\mathrm{cm}\) ). What magnifications can be obtained with the microscope if it is adjusted so that the image formed by the objective is \(17 \mathrm{~cm}\) from that lens? Because \(s_{i o}=17 \mathrm{~cm}\) the magnification formula for a microscope, with \(d_{n}=25 \mathrm{~cm}\), gives the following possibilities for \(M_{A}\) : For \(f_{E}=3 \mathrm{~cm}, f_{o}=0.3 \mathrm{~cm}:\) For \(f_{E}=3 \mathrm{~cm}, f_{o}=0.7 \mathrm{~cm}:\) For \(f_{g}=5 \mathrm{~cm}, f_{\theta}=0.3 \mathrm{~cm}:\) For \(f_{E}=5 \mathrm{~cm}, f_{o}=0.7 \mathrm{~cm}:\) \(M_{A}=(9.33)(55,6)=518=5.2 \times 10^{2}\) \(M_{A}=(9.33)(23.2)=216=2.2 \times 10^{2}\) \(M_{A}=(5)(55.6)=278=2.8 \times 10^{2}\) \(M_{A}=(5)(23.2)=116=1.2 \times 10^{2}\)

In the compound microscope shown, the objective and eyepiece have focal lengths of \(+0.80\) and \(+2.5 \mathrm{~cm}\), respectively. The real intermediate image \(A^{\prime} B^{\prime}\) formed by the objective is \(16 \mathrm{~cm}\) from the objective. Determine the total magnification if the eye is held close to the eyepiece and views the virtual image \(A^{\prime \prime} B^{\prime \prime}\) at a distance of \(25 \mathrm{~cm}\). Method 1 Let \(s_{o} 0=\) Object distance from the objective \(s_{o} 0=\) Real-image distance from the objective $$ \frac{1}{s_{o o}}=\frac{1}{f_{o}}-\frac{1}{s_{i O}}=\frac{1}{0.80}-\frac{1}{16}=\frac{19}{16} \mathrm{~cm}^{-1} $$ and so the objective produces the linear magnification $$ M_{T O}=-\frac{s_{i O}}{s_{o o}}=-(16 \mathrm{~cm})=\left(\frac{19}{16} \mathrm{~cm}^{-1}\right)=-19 $$ The intermediate image is inverted. The magnifying power of the eyepiece is $$ M_{T E}=-\frac{s_{i E}}{s_{o E}}=-s_{i E}\left(\frac{1}{f_{E}}-\frac{1}{s_{i E}}\right)=-\frac{s_{i E}}{f_{E}}+1=-\frac{-25}{+2.5}+1=11 $$ The eyepiece does not flip the image: the intermediate image is inverted and the final image is inverted. Therefore, the magnifying power of the instrument is \(-19 \times 11=-2.1 \times 10^{2}\). Alternatively, under the conditions stated, the magnifying power of the eyepiece can be found as $$ \frac{25}{f_{E}}+1=\frac{25}{2.5}+1=11 $$ Method 2 From Eq. (39.2) with \(s_{i o}=16 \mathrm{~cm}\), $$ \text { Magnification }=\left(\frac{d_{n}}{f_{E}}+1\right)\left(\frac{s_{i o}}{f_{o}}-1\right)=\left(\frac{25}{2.5}+1\right)\left(\frac{16}{0.8}-1\right)=2.1 \times 10^{2} $$

Two lenses, of focal lengths \(+6.0 \mathrm{~cm}\) and \(-10 \mathrm{~cm}\), are spaced \(1.5\) \(\mathrm{cm}\) apart. Locate and describe the image of an object \(30 \mathrm{~cm}\) in front of the \(+6.0\) -cm lens.

An engraver who has normal eyesight uses a converging lens of focal length \(8.0 \mathrm{~cm}\), which he holds very close to his eye. At what distance from the work should the lens be placed, and what is the magnification of the lens? Method 1 When a converging lens is used as a magnifying glass, the object is between the lens and the focal point. The virtual erect, and enlarged image forms at the distance of distinct vision, \(25 \mathrm{~cm}\) from the eye. For a virtual image \(s_{i}<0\). Thus, Method 2 By the formula, $$ M_{A}=\frac{d_{n}}{f}+1=\frac{25}{8.0}+1=4.1 $$ Note that in this simple case \(M_{T}=M_{A}\).
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