The objective lens in a laboratory microscope has a focal length of \(3.00 \mathrm{~cm}\) and provides an overall magnification of \(1.0 \cdot 10^{2} .\) What is the focal length of the eyepiece if the distance between the two lenses is \(30.0 \mathrm{~cm}\) ?

Short Answer

Expert verified
Answer: The focal length of the eyepiece lens is 6.00 cm.

Step by step solution

01

Identify the given values.

The problem provides us with the following information: - Focal length of the objective lens \((f_o) = 3.00 \mathrm{~cm}\) - Overall magnification \((M) = 1.0 \cdot 10^2\) - Distance between the two lenses \((L) = 30.0 \mathrm{~cm}\)
02

Write down the lens magnification formula.

The magnification formula for lenses is given by: \(M = M_o \cdot M_e\) Where \(M_o\) is the magnification produced by the objective lens and \(M_e\) is the magnification produced by the eyepiece lens.
03

Write down the formula relating focal length, object distance, and image distance.

The lens formula relating focal length, object distance \((d_o)\), and image distance \((d_i)\) is: \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\)
04

Write down the formula relating magnification, object distance, and image distance.

The formula relating magnification, object distance \((d_o)\), and image distance \((d_i)\) is: \(M = \frac{d_i}{d_o}\)
05

Calculate the magnification produced by the objective lens.

We will use the magnification formula to solve for \(M_o\) using the given overall magnification \((M)\) and the fact that the eyepiece is at the normal adjustment position, which means that the angular magnification of the eyepiece is 5.00 (this is a general rule for microscope eyepieces). So, \(M_o = \frac{M}{M_e} = \frac{1.0 \cdot 10^2}{5.00} = 20\)
06

Calculate the image distance for the objective lens.

We will use the magnification formula for the objective lens and the given object distance to calculate the image distance \((d_{io})\): \(M_o = \frac{d_{io}}{d_{oo}}\) We will assume that the object is very close to the objective lens, so \(d_{oo} \approx f_o\). Solving for \(d_{io}\): \(d_{io} = M_o \cdot d_{oo} = 20 \cdot 3.00 = 60.0 \mathrm{~cm}\)
07

Calculate the object distance for the eyepiece lens.

Since the distance between the two lenses \((L)\) is given as \(30.0 \mathrm{~cm}\), the object distance for the eyepiece lens \((d_{oe})\) is: \(d_{oe} = d_{io} - L = 60.0 \mathrm{~cm} - 30.0 \mathrm{~cm} = 30.0 \mathrm{~cm}\)
08

Calculate the image distance for the eyepiece lens.

As we know the angular magnification for the eyepiece lens being \(M_e = 5.00\), we can calculate the image distance for the eyepiece lens \((d_{ie})\) using the magnification formula: \(d_{ie} = M_e \cdot d_{oe} = 5.00 \cdot 30.0 = 150.0 \mathrm{~cm}\)
09

Calculate the focal length of the eyepiece lens.

Now, using the lens formula for the eyepiece lens, we can determine its focal length \((f_e)\): \(\frac{1}{f_e} = \frac{1}{d_{oe}} + \frac{1}{d_{ie}}\) Solve for \(f_e\): \(f_e = \frac{1}{\frac{1}{d_{oe}} + \frac{1}{d_{ie}}} = \frac{1}{\frac{1}{30.0} + \frac{1}{150.0}} = \frac{1}{\frac{1}{6}} = 6.00 \mathrm{~cm}\) The focal length of the eyepiece lens is \(6.00 \mathrm{~cm}\).

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Key Concepts

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

Lens Magnification Formula
Understanding the concept of magnification in microscopes is crucial for students studying optics. The lens magnification formula connects the size of the image seen through the microscope to the actual size of the object being viewed.

It is expressed as: \(M = M_o \times M_e\), where \(M\) is the total magnification, \(M_o\) is the magnification by the objective lens, and \(M_e\) is the magnification by the eyepiece lens. When dealing with microscopy, it's important to note that the overall magnification is a product of both the objective lens and the eyepiece lens. Thus, to find the magnification of one lens, you can divide the total magnification by the magnification of the other lens, as we demonstrated in the original exercise solution.
Focal Length
The focal length of a lens is a key concept in optics that describes how strongly it converges or diverges light. It is the distance from the lens to the focal point, where rays of light that are parallel to the optical axis will converge after passing through the lens.

In simple terms, a shorter focal length means the lens is more powerful, as it bends the light rays more sharply, bringing them to focus in a shorter distance. The focal length is critical in determining the magnifying power of the lens—in microscopes, a shorter focal length for the objective lens usually implies a higher magnification power. For eyepieces, a longer focal length provides comfort but decreases magnification power; hence the careful balance necessary in the design of optical instruments like microscopes.
Image Distance
The image distance (\(d_i\)) is the distance from the lens to the image formed by that lens. It's one of the variables in the lens formula \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\).

When calculating the image distance, you're essentially determining where the image will appear with respect to the lens' position. This concept is directly linked to the creation of magnified images in microscopes. A greater image distance means that the lens creates a larger image further away. In microscopic calculations, this distance plays a pivotal role in finding other related optical values such as magnification, as seen in our exercise.
Object Distance
The object distance (\(d_o\)) refers to the distance between the object and the lens. It is an integral part of the lens formula, alongside the focal length and the image distance. In the context of microscopes, the object distance is specifically the distance between the objective lens and the specimen being viewed.

This measure is used to calculate the magnification and is usually set at a fixed length in microscopes, known as the tube length. Adjusting the object distance allows the user to focus on the specimen. When the object is placed at the focal length of the objective lens, it results in a magnified and inverted image at infinity, which is often the basis for calculations involving the eyepiece lens.

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Most popular questions from this chapter

Where is the image formed if an object is placed \(25 \mathrm{~cm}\) from the eye of a nearsighted person. What kind of a corrective lens should the person wear? a) Behind the retina. Converging lenses. b) Behind the retina. Diverging lenses. c) In front of the retina. Converging lenses. d) In front of the retina. Diverging lenses.

Two refracting telescopes are used to look at craters on the Moon. The objective focal length of both telescopes is \(95.0 \mathrm{~cm}\) and the eyepiece focal length of both telescopes is \(3.80 \mathrm{~cm} .\) The telescopes are identical except for the diameter of the lenses. Telescope A has an objective diameter of \(10.0 \mathrm{~cm}\) while the lenses of telescope \(\mathrm{B}\) are scaled up by a factor of two, so that its objective diameter is \(20.0 \mathrm{~cm}\). a) What are the angular magnifications of telescopes \(A\) and \(B\) ? b) Do the images produced by the telescopes have the same brightness? If not, which is brighter and by how much?

A plastic cylinder of length \(3.0 \cdot 10^{1} \mathrm{~cm}\) has its ends ground to convex (from the rod outward) spherical surfaces, each having radius of curvature \(1.0 \cdot 10^{1} \mathrm{~cm}\). A small object is placed \(1.0 \cdot 10^{1} \mathrm{~cm}\) from the left end. How far will the image of the object lie from the right end, if the index of refraction of the plastic is \(1.5 ?\)

Three converging lenses of focal length \(5.0 \mathrm{~cm}\) are arranged with a spacing of \(2.0 \cdot 10^{1} \mathrm{~cm}\) between them, and are used to image an insect \(2.0 \cdot 10^{1} \mathrm{~cm}\) away. a) Where is the image? b) Is it real or virtual? c) Is it upright or inverted?

Mirrors for astronomical instruments are invariably first-surface mirrors: The reflective coating is applied on the surface exposed to the incoming light. Household mirrors, on the other hand, are second-surface mirrors: The coating is applied to the back of the glass or plastic material of the mirror. (You can tell the difference by bringing the tip of an object close to the surface of the mirror. Object and image will nearly touch with a first-surface mirror; a gap will remain between them with a second-surface mirror.) Explain the reasons for these design differences.

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