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Resolution of a Microscope. The image formed by a microscope objective with a focal length of 5.00 \(\mathrm{mm}\) is 160 \(\mathrm{mm}\) from its second focal point. The eyepiece has a focal length of 26.0 \(\mathrm{mm}\) . (a) What is the angular magnification of the microscope? (b) The unaided eye can distinguish two points at its near point as separate if they are about 0.10 \(\mathrm{mm}\) apart. What is the minimum separation between two points that can be observed (or resolved) through this microscope?

Short Answer

Expert verified
(a) 308.44. (b) 0.000324 mm.

Step by step solution

01

Understand the Problem

We are given a microscope with an objective of focal length 5 mm and an image that is formed 160 mm from the second focal point of the objective lens. The eyepiece has a focal length of 26 mm. We need to find the angular magnification and the resolution of the microscope.
02

Calculate the Magnification of the Objective Lens

The magnification of the objective lens (M_o) is given by the formula:\[ M_o = \frac{L}{f_o} \]where L is the tube length (160 mm) and f_o is the focal length of the objective lens (5 mm). Thus,\[ M_o = \frac{160}{5} = 32 \]
03

Calculate the Angular Magnification of the Eyepiece

The angular magnification of the eyepiece (M_e) is given by:\[ M_e = \frac{25}{f_e} \]where 25 cm is the near point of the average human eye (conventional value) and f_e is the focal length of the eyepiece (26 mm or 2.6 cm). Thus,\[ M_e = \frac{25}{2.6} \approx 9.62 \]
04

Calculate the Total Angular Magnification of the Microscope

The total angular magnification (M_t) of the microscope is the product of the magnifications of the objective lens and eyepiece:\[ M_t = M_o \cdot M_e = 32 \cdot 9.62 \approx 308.44 \]
05

Determine the Resolution of the Microscope

The resolution of the microscope, or the minimum separation (d_{min}) that can be observed, is given by dividing the smallest resolvable distance by the total magnification:\[ d_{min} = \frac{d_{eye}}{M_t} \]where d_{eye} is the minimum separation the unaided eye can resolve at the near point (0.10 mm). Thus,\[ d_{min} = \frac{0.10}{308.44} \approx 0.000324 \text{ mm} \]
06

Conclusion

(a) The angular magnification of the microscope is approximately 308.44. (b) The minimum separation that can be resolved with the microscope is approximately 0.000324 mm.

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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 through an optical device compared to how it appears to the naked eye. It is especially important in microscopes because it helps us see tiny details clearly. In a microscope, total angular magnification is the product of the magnification provided by the objective lens and the eyepiece. For this specific example, the total angular magnification was calculated to be approximately 308.44. This means that the microscope makes an object appear 308 times larger than it would to the bare eye.
To break it down further:
  • The objective lens magnifies the image first.
  • The eyepiece then magnifies this already magnified image, giving us a much closer view.
This large magnification is crucial in scientific studies where observing minute details at higher magnification levels is necessary.
Objective Lens Magnification
The objective lens is a critical component in a microscope as it provides the primary magnification. In this case, the objective lens magnification was determined using the formula: \[ M_o = \frac{L}{f_o} \] where
  • \(L\) is the length of the tube, which was given as 160 mm, and
  • \(f_o\) is the focal length of the objective lens, provided as 5 mm.
Plugging these values into the formula, the magnification of the objective lens is calculated to be 32. This means the objective lens alone makes the object appear 32 times larger than it actually is.
Understanding this concept helps in grasping how different parts of a microscope contribute to the total power to enlarge the image.
Eyepiece Magnification
The eyepiece of a microscope further magnifies the image formed by the objective lens. The angular magnification of the eyepiece is evaluated using the formula: \[ M_e = \frac{25}{f_e} \] where
  • 25 cm (or 250 mm) is the near point of a normal human eye, and
  • \(f_e\) is the focal length of the eyepiece, which in this exercise is 26 mm (or 2.6 cm).
After calculation, the eyepiece magnifies the image about 9.62 times.
The eyepiece acts as a magnifying glass for the already enlarged image formed by the objective lens, contributing significantly to the overall magnification power of the microscope. This enhances the user's ability to see the fine details of small specimens.
Minimum Separation Resolution
Resolution is a measure of the microscope's ability to distinguish two close points as separate. This is critical in applications where detail is essential, such as cell biology. The minimum separation resolution tells us the smallest distance between two points that can be distinguished by the microscope.
The formula used in this context is: \[ d_{min} = \frac{d_{eye}}{M_t} \] where
  • \(d_{eye}\) is the smallest separation the eye can distinguish unaided (0.10 mm in this exercise), and
  • \(M_t\) is the total angular magnification (308.44 for this example).
With these values, the minimum separation resolution calculates to approximately 0.000324 mm.
This means that through the microscope, we can see details as close as 0.000324 mm apart, showcasing the power and precision of the microscope in enhancing our visual capabilities.

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Most popular questions from this chapter

A microscope with an objective of focal length 8.00 \(\mathrm{mm}\) and an eyepiece of focal length 7.50 \(\mathrm{cm}\) is used to project an image on a screen 2.00 \(\mathrm{m}\) from the eyepiece. Let the image distance of the objective be 18.0 \(\mathrm{cm}\) . (a) What is the lateral magnification of the image? (b) What is the distance between the objective and the eyepiece?

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 film be changed? To refocus on the more distant object, is the lens moved toward or away from the film?

Choosing a Camera Lens. The picture size on ordinary \(35-\mathrm{mm}\) camera film is \(24 \mathrm{mm} \times 36 \mathrm{mm}\) . Focal lengths of lenses available for \(35-\mathrm{mm}\) cameras typically include \(28,35,50\) (the "normal" lens), \(85,100,135,200,\) and \(300 \mathrm{mm},\) among others. Which of these lenses should be used to photograph the following objects, assuming that the object is to fill most of the picture area? (a) a building 240 \(\mathrm{m}\) tall and 160 \(\mathrm{m}\) wide at a distance of 600 \(\mathrm{m}\) , and (b) a mobile home 9.6 \(\mathrm{m}\) in length at a distance of 40.0 \(\mathrm{m}\) .

You hold a spherical salad bowl 90 \(\mathrm{cm}\) in front of your face with the bottom of the bowl facing you. The salad bowl is made of polished metal with a \(35-\mathrm{cm}\) radius of curvature. (a) Where is the image of your 2.0 -cm-tall nose located? (b) What are the image's size, orientation, and nature (real or virtual)?

A spherical, concave, shaving mirror has a radius of curvature of \(32.0 \mathrm{cm} .\) (a) What is the magnification of a person's face when it is 12.0 \(\mathrm{cm}\) to the left of the vertex of the mirror? (b) Where is the image? Is the image real or virual? (c) Draw a principal-ray diagram showing the formation of the image.

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