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Chapter 25: Problem 38

\(\bullet\) 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 that can be resolved with this microscope?

Short Answer

Expert verified
(a) 339.68, (b) 0.000294 mm.

Step by step solution

01

Identify Given Values

The microscope objective has a focal length of \( f_o = 5.00 \, \text{mm} \) and the image is formed \( d_i = 160 \, \text{mm} \) from its second focal point. The eyepiece has a focal length of \( f_e = 26.0 \, \text{mm} \).
02

Calculate Objective Lens Magnification

The magnification of the objective lens, \( M_o \), is given by \[M_o = \frac{d_i}{f_o} = \frac{160 \, \text{mm}}{5.00 \, \text{mm}} = 32.\]
03

Calculate Eyepiece Angular Magnification

The angular magnification of the eyepiece, \( M_e \), is calculated using the formula \[M_e = 1 + \frac{25 \, \text{cm}}{f_e} = 1 + \frac{25.0 \, \text{cm}}{2.6 \, \text{cm}} = 10.615.\]
04

Calculate Total Angular Magnification

The total angular magnification of the microscope \( M \) is the product of the magnifications of the objective and eyepiece: \[M = M_o \times M_e = 32 \times 10.615 = 339.68.\]
05

Minimum Resolvable Separation With Microscope

The minimum separation \( d \) that can be resolved is determined by dividing the unaided minimum separation by the total magnification:\[d = \frac{0.10 \, \text{mm}}{M} = \frac{0.10 \, \text{mm}}{339.68} \approx 2.94 \times 10^{-4} \, \text{mm} \approx 0.000294 \, \text{mm}.\]

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Key Concepts

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

Angular Magnification
Angular magnification in a microscope is a crucial factor that defines how much larger an object appears compared to its actual size. This is the combined effect of both the objective lens and the eyepiece working together. It involves:
  • Objective lens magnification: This occurs when the objective lens creates an enlarged image of the specimen. The formula used is \( M_o = \frac{d_i}{f_o} \), where \( d_i \) is the image distance from the lens and \( f_o \) is the focal length.
  • Eyepiece magnification: This happens when the eyepiece further enlarges the image produced by the objective lens. The formula, \( M_e = 1 + \frac{25 \, \text{cm}}{f_e} \), takes into consideration the viewer's near point (usually 25 cm for a normal eye).
The overall angular magnification \( M \) is the product of these two magnifications \( M = M_o \times M_e \). This exponential increase in size allows scientists to view minute details that are otherwise invisible to the naked eye.
Focal Length
Focal length is a fundamental property of lenses that influences how they focus light. For microscopes, it determines how lenses bend light rays:
  • The objective lens has a short focal length (e.g., 5.00 mm in this case), meaning it can produce a much larger image for the same size specimen when placed close to the object.
  • The eyepiece generally has a longer focal length (e.g., 26.0 mm), which helps further magnify the image formed by the objective lens.
Understanding the focal length of both lenses helps in calculating the individual and total magnifications. The focal length determines where the image is formed in relation to the lens, influencing the clarity and size of what you see through the microscope. Careful calibration of these lengths ensures improved resolution and detail.
Objective Lens
The objective lens is the primary lens in microscopes that is closest to the specimen. Its main function is to:
  • Gather light from the specimen.
  • Produce a real, inverted image that is magnified.
The strength of the objective lens is defined by its focal length and magnification power. In the given exercise, the objective lens had a focal length of 5.00 mm, producing a high level of magnification by creating an intermediate image far larger than the specimen itself. This lens is vital because much of the microscope's total magnification comes from here. Choosing the right objective lens can drastically affect the clarity and detail of microscopic observations.
Eyepiece
The eyepiece, also known as the ocular lens, is placed at the top of a microscope and is the final lens that the observer looks through. Its purposes include:
  • Further magnifying the real image produced by the objective lens.
  • Making the image visible to the human eye by transforming the light rays into a virtual image.
Generally, eyepieces have a lower magnification than objective lenses. In the exercise, the eyepiece had a focal length of 26.0 mm. By enlarging the image received from the objective lens, it adjusts the field of view and aids in comfortable observation. Overall, both the objective lens and eyepiece work synergistically. While the objective lens provides the bulk of the magnification, the eyepiece fine-tunes it for optimal viewing, allowing users to explore the tiny details of their specimens effectively.

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Most popular questions from this chapter

. A 135 mm telephoto lens for a 35 mm camera has \(f\) -stops that range from \(f / 2.8\) to \(f / 22\) . (a) What are the smallest and largest aperture diameters for this lens? What is the diameter at \(f / 11 ?\) (b) If a 50 \(\mathrm{mm}\) lens had the same \(f-\) stops as the telephoto lens, what would be the smallest and largest aperture diameters for that lens? (c) At a given shutter speed, what is the ratio of the greatest to the smallest light intensity of the film image? (d) If the shutter speed for correct exposure at \(f / 22\) is 1\(/ 30\) s, what shutter speed is needed at \(f / 2.8 ?\)

A camera lens has a focal length of 200 \(\mathrm{mm}\) . How far from the lens should the subject for the photo be if the lens is 20.4 \(\mathrm{cm}\) from the film?

\bullet 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 student's far point is at \(17.0 \mathrm{cm},\) and she needs glasses to view her computer screen comfortably at a distance of 45.0 \(\mathrm{cm} .\) What should be the power of the lenses for her glasses?

Contact lenses. 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 nearson nearsighted or farsighted? (b) What type of lens (con- verging 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?
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