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Chapter 33: Problem 78

(II) The eyepiece of a compound microscope has a focal length of \(2.80 \mathrm{~cm}\) and the objective lens has \(f=0.740 \mathrm{~cm} .\) If an object is placed \(0.790 \mathrm{~cm}\) from the objective lens, calculate (a) the distance between the lenses when the microscope is adjusted for a relaxed eye, and ( \(b\) ) the total magnification.

Short Answer

Expert verified
(a) Distance between lenses: sum of \( d_{io} \) and \( f_e \). (b) Total magnification: \( M_o \times M_e \).

Step by step solution

01

Understand the Problem

We are given a compound microscope with details of the eyepiece and objective lens. We need to find the lens separation for a relaxed eye and the total magnification.
02

Calculate the Image Distance from the Objective Lens

For the objective lens, which forms an image of the object, we use the lens formula: \[ \frac{1}{f_o} = \frac{1}{d_o} + \frac{1}{d_{io}} \] where \( f_o = 0.740 \) cm, and \( d_o = 0.790 \) cm. Solving for \( d_{io} \) (the image distance from the objective lens):\[ \frac{1}{0.740} = \frac{1}{0.790} + \frac{1}{d_{io}} \]\[ \frac{1}{d_{io}} = \frac{1}{0.740} - \frac{1}{0.790} \]Calculate \( d_{io} \).
03

Determine the Image Distance from the Eyepiece

For a relaxed eye, the final image is at infinity. Thus, the image formed by the eyepiece is at its focal point, \( f_e = 2.80 \) cm. Hence, the distance of the image formed by the objective lens is the sum of \( d_{io} \) and the focal length of the eyepiece \( f_e \).
04

Calculate the Distance Between the Lenses

Using the results from Steps 2 and 3, we calculate the distance between the two lenses: \( \text{Distance between lenses} = d_{io} + f_e \).
05

Calculate the Magnification of the Objective Lens

The magnification of the objective lens \( M_o \) is given by: \[ M_o = \frac{d_{io}}{d_o} \]. Use \( d_{io} \) obtained in Step 2 and \( d_o = 0.790 \) cm.
06

Calculate the Magnification of the Eyepiece

The magnification of the eyepiece (assuming the final image is at infinity for a relaxed eye) is \( M_e = \frac{25}{f_e} \) where \( f_e = 2.80 \) cm and 25 cm is the assumed near point distance conventionally used.
07

Determine the Total Magnification

The total magnification \( M_t \) of the microscope is the product of the magnifications of the objective and the eyepiece: \[ M_t = M_o \times M_e \]. Substitute the values found in Steps 5 and 6 to find \( M_t \).

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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 the distance from the lens where a beam of light meets after passing through. It's a crucial factor when determining how a lens will focus light rays. In the case of a compound microscope, different lenses contribute to the overall image creation process.

The objective lens and eyepiece both have specific focal lengths. For example, in the given exercise, the focal length of the eyepiece is given as 2.80 cm, while the objective lens has a much shorter focal length of 0.740 cm. These values influence how the light is bent, focused, and thus how the final image is formed and magnified.
Magnification
Magnification is one of the primary advantages of using a microscope, as it enables us to see tiny details of small objects. It is the process of enlarging the apparent size, not the physical size, of something.

In a compound microscope, magnification occurs in two main stages: through the objective lens and through the eyepiece. The objective lens creates an enlarged image of the object, and the eyepiece further magnifies this image. The total magnification is the product of these two individual magnifications, and in practical terms, it can allow you to see objects at a much larger scale, revealing details not visible to the naked eye.
Lens Formula
The lens formula is critical for determining the relationship between the object distance, image distance, and focal length of a lens. It is given by the equation: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]

In this formula, \(f\) represents the focal length, \(d_o\) is the object distance from the lens, and \(d_i\) is the image distance from the lens.

By applying this formula, we can solve for one of these variables if the other two are known, which helps in determining how and where the image will form when using multiple lenses in a compound microscope.
Objective Lens
The objective lens is one of the two main lenses in a compound microscope and is crucial in initial image formation. It is located close to the specimen and is responsible for creating a magnified image of the specimen.

The optical power of the objective lens is characterized by its short focal length, as seen in the exercise where it is 0.740 cm. This short focal length allows it to produce a large amount of initial magnification, setting the stage for further magnification by the eyepiece.
Eyepiece
The eyepiece, also known as the ocular lens, is where you place your eye to observe the image. It further magnifies the image formed by the objective lens, resulting in the final magnified view that you see.

Its magnification power is determined largely by its focal length. The shorter the focal length, the higher the magnification. For example, with a focal length of 2.80 cm as in the exercise, a typical eyepiece can add significant magnification to the initial image created by the objective lens. Moreover, when the microscope is adjusted for a relaxed eye, the eyepiece focuses the image at infinity, allowing for comfortable viewing over longer periods.

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Most popular questions from this chapter

(II) An inexpensive instructional lab microscope allows the user to select its objective lens to have a focal length of \(32 \mathrm{~mm}, 15 \mathrm{~mm},\) or \(3.9 \mathrm{~mm} .\) It also has two possible eyepieces with magnifications \(5 \times\) and \(10 \times\). Each objective forms a real image \(160 \mathrm{~mm}\) beyond its focal point. What are the largest and smallest overall magnifications obtainable with this instrument?

(II) A \(-8.00-\mathrm{D}\) lens is held \(12.5 \mathrm{~cm}\) from an ant \(1.00 \mathrm{~mm}\) high. Describe the position, type, and height of the image.

(II) A person struggles to read by holding a book at arm's length, a distance of \(55 \mathrm{~cm}\) away. What power of reading glasses should be prescribed for her, assuming they will be placed \(2.0 \mathrm{~cm}\) from the eye and she wants to read at the "normal" near point of \(25 \mathrm{~cm} ?\)

(III) In the "magnification" method, the focal length \(f\) of a converging lens is found by placing an object of known size at various locations in front of the lens and measuring the resulting real-image distances \(d_{i}\) and their associated magnifications \(m\) (minus sign indicates that image is inverted). The data taken in such an experiment are given here:\(\begin{array}{rrrrrr}{d_{\mathrm{i}}(\mathrm{cm})} & {20} & {25} & {30} & {35} & {40} \\ {m} & {-0.43} & {-0.79} & {-1.14} & {-1.50} & {-1.89}\end{array}\) (a) Show analytically that a graph of \(m\) vs. \(d_{\text { i should }}\) produce a straight line. What are the theoretically expected values for the slope and the \(y\) -intercept of this line? [Hint: \(d_{\mathrm{o}}\) is not constant.] \((b)\) Using the data above, graph \(m\) vs. \(d_{\mathrm{i}}\) and show that a straight line does indeed result. Use the slope of this line to determine the focal length of the lens. Does the \(y\) -intercept of your plot have the expected value? (c) In performing such an experiment, one has the practical problem of locating the exact center of the lens since \(d_{\mathrm{i}}\) must be measured from this point. Imagine, instead, that one measures the image distance \(d\) from the back surface of the lens, which is a distance \(\ell\) from the lens's center. Then, \(d_{i}=d_{1}^{\prime}+\ell .\) Show that, when implementing the magnification method in this fashion, a plot of \(m\) vs.di will still result in a straight line. How can \(f\) be determined from this straight line?

(II) Show analytically that the image formed by a converging lens \((a)\) is real and inverted if the object is beyond the focal point \(\left(d_{0}>f\right),\) and \((b)\) is virtual and upright if the object is within the focal point \(\left(d_{\mathrm{o}}f,\) and \((d)\) for \(0<-d_{0}
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