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Chapter 34: Problem 54

A certain microscope is provided with objectives that have focal lengths of \(16 \mathrm{mm}, 4 \mathrm{mm},\) and 1.9 \(\mathrm{mm}\) and with eyepieces that have angular magnifications of \(5 \times\) and \(10 \times .\) Each objective forms an image 120 \(\mathrm{mm}\) beyond its second focal point. Determine (a) the largest overall angular magnification obtainable and (b) the least overall angular magnification obtainable.

Short Answer

Expert verified
Largest magnification: 641.6 and smallest magnification: 42.5.

Step by step solution

01

Understand the Problem

We need to find the largest and smallest overall angular magnifications obtainable using the microscope's objectives and eyepieces. Angular magnification of a microscope is given by the product of the magnification of the objective and the eyepiece.
02

Identify Key Formulas and Values

The objective lens forms an image a distance of 120 mm beyond its second focal point. The magnification of the objective, \( M_o \), is given by \( M_o = \frac{v}{f_o} \), where \( v = 120 \, \text{mm} + f_o \) (since the objective forms an image 120 mm beyond its focal length \( f_o \)). The overall magnification is \( M = M_o \times M_e \), where \( M_e \) is the magnification of the eyepiece (5 or 10).
03

Calculate Objective Magnifications

For each objective: - For the 16 mm objective: \( v = 120 + 16 = 136 \, \text{mm} \) and \( M_o = \frac{136}{16} = 8.5 \).- For the 4 mm objective: \( v = 120 + 4 = 124 \, \text{mm} \) and \( M_o = \frac{124}{4} = 31 \).- For the 1.9 mm objective: \( v = 120 + 1.9 = 121.9 \, \text{mm} \) and \( M_o = \frac{121.9}{1.9} \approx 64.16 \).
04

Determine Largest Overall Angular Magnification

We choose the objective with the highest magnification (1.9 mm, \( M_o \approx 64.16 \)) and the eyepiece with the highest magnification (\( M_e = 10 \)). The largest overall magnification is thus \( M = 64.16 \times 10 = 641.6 \).
05

Determine Least Overall Angular Magnification

We choose the objective with the lowest magnification (16 mm, \( M_o = 8.5 \)) and the eyepiece with the lowest magnification (\( M_e = 5 \)). The smallest overall magnification is thus \( M = 8.5 \times 5 = 42.5 \).

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Key Concepts

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

Objective Lenses
Objective lenses are an essential part of a microscope and significantly contribute to the magnification process. They are the lenses closest to the specimen, and their primary role is to create an enlarged image of the object.

Every objective lens has a specific focal length, which is the distance from the lens to the point where parallel light rays converge to a single point. For instance, in the given microscope, we have objective lenses with focal lengths of 16 mm, 4 mm, and 1.9 mm.
  • The shorter the focal length, the higher the magnification capability of the lens.
  • This is because a shorter focal length brings the light rays closer together, creating a larger image.
  • Therefore, the 1.9 mm lens provides more magnification than the 16 mm lens.
In practice, users select the lens with the desired focal length to achieve the necessary magnification for their observations.
Eyepiece Magnification
The eyepiece, also known as the ocular lens, is what you look through at the top of the microscope. Its purpose is to magnify the image formed by the objective lens.

The microscope in the exercise has eyepieces with magnifications of 5脳 and 10脳. This means that the eyepiece will enlarge the image by five or ten times its original size, respectively.
  • Choosing a higher magnification eyepiece increases the overall magnification of the microscope.
  • However, higher magnification can sometimes lead to a less clear image, as it may magnify the imperfections or limits of the objective's image.
Thus, while using an eyepiece, the balance between magnification and clarity needs to be considered for effective observations.
Focal Length
Focal length is critical in determining the magnification of an objective lens within a microscope. It's essentially the lens's optical strength, indicating how strongly the lens converges or diverges light rays.

In calculating magnification, the formula for the objective lens's magnification is given as:
\[ M_o = \frac{v}{f_o} \]where:
  • \( v \) is the image distance, the distance from the lens where the image is formed, which in this case is 120 mm beyond the focal length.
  • \( f_o \) is the lens's focal length.
Through this calculation, we understand that as the focal length decreases, magnification inherently increases, which explains why the 1.9 mm lens yields a higher magnification than the 16 mm lens.
Overall Magnification Calculation
The overall magnification of a microscope is a product of the objective lens's magnification and the eyepiece magnification. This is a fundamental principle in understanding how microscopes enlarge images.

The formula is given as:
\[ M = M_o \times M_e \]where:
  • \( M \) is the overall magnification.
  • \( M_o \) is the magnification of the objective lens, which depends on its focal length.
  • \( M_e \) is the magnification of the eyepiece.
To find the largest magnification, combine the objective lens with the highest magnification (\(1.9 \, \text{mm} \) focal length) and the eyepiece with the highest magnification (10脳). To find the smallest, use the lens with the longest focal length (16 mm) paired with the 5脳 eyepiece. Hence, the overall magnification ranges from 42.5脳 to 641.6脳 in the given exercise.

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Most popular questions from this chapter

Two thin lenses with focal lengths of magnitude \(15.0 \mathrm{cm},\) the first diverging and the second converging, are placed 12.00 \(\mathrm{cm}\) apart. An object 4.00 \(\mathrm{mm}\) tall is placed 5.00 \(\mathrm{cm}\) to the left of the first (diverging) lens. (a) Where is the image formed by the first lens located? (b) How far from the object is the final image formed?(c) Is the final image real or virtual? (d) What is the height of the final image? Is the final image erect or inverted?

Three thin lenses, each with a focal length of 40.0 cm, are aligned on a common axis; adjacent lenses are separated by 52.0 \(\mathrm{cm}\) . Find the position of the image of a small object on the axis, 80.0 \(\mathrm{cm}\) to the left of the first lens.

A thin lens with a focal length of 6.00 \(\mathrm{cm}\) is used as a simple magnifier. (a) What angular magnification is obtainable with the lens if the object is at the focal point? (b) When an object is examined through the lens, how close can it be brought to the lens? Assume that the image viewed by the eye is at the near point, 25.0 \(\mathrm{cm}\) from the eye, and that the lens is very close to the eye.

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.

A solid glass hemisphere of radius 12.0 \(\mathrm{cm}\) and index of refraction \(n=1.50\) is placed with its flat face downward on a table. A parallel beam of light with a circular cross section 3.80 \(\mathrm{mm}\) in diameter travels straight down and enters the hemisphere at the center of its curved surface. (a) What is the diameter of the circle of light formed on the table? (b) How does your result depend on the radius of the hemisphere?
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