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Chapter 24: Problem 38

Jordan is building a compound microscope using an eyepiece with a focal length of \(7.50 \mathrm{cm}\) and an objective with a focal length of $1.500 \mathrm{cm} .\( He will place the specimen a distance of \)1.600 \mathrm{cm}$ from the objective. (a) How far apart should Jordan place the lenses? (b) What will be the angular magnification of this microscope?

Short Answer

Expert verified
Answer: Jordan should place the lenses 4.5 cm apart, and the approximate angular magnification of the microscope is 9.75 times.

Step by step solution

01

Note down the given values

We have the following values: Focal length of the eyepiece: \(f_e = 7.5 \mathrm{cm}\) Focal length of the objective: \(f_o = 1.5 \mathrm{cm}\) Distance from the specimen to the objective: \(s_o = 1.6 \mathrm{cm}\)
02

Find the image distance for the objective lens

We can use the thin lens equation: \(\frac{1}{f_o} = \frac{1}{s_o} + \frac{1}{s_i^o}\) We can solve this equation for the image distance \(s_i^o\): \(s_i^o = \frac{1}{\frac{1}{f_o} - \frac{1}{s_o}}\) Now, plug in the values of \(f_o\) and \(s_o\): \(s_i^o = \frac{1}{\frac{1}{1.5} - \frac{1}{1.6}} \approx 12.0 \mathrm{cm}\)
03

Calculate the magnification for the objective lens

We can use the magnification formula for a single lens: \(M_o = \frac{s_i^o}{s_o}\) Now, plug in the values of \(s_i^o\) and \(s_o\): \(M_o = \frac{12.0}{1.6} = 7.5\)
04

Determine the location for the final image using the eyepiece lens

The final image will be at the focus of the eyepiece, so the image distance for the eyepiece lens is equal to its focal length: \(s_i^e = f_e = 7.5 \mathrm{cm}\)
05

Determine the distance between the lenses (objective and eyepiece)

To find the distance between the objective and eyepiece, we'll calculate the object distance for the eyepiece lens: \(s_o^e = s_i^o - s_i^e\) Now, plug in the values of \(s_i^o\) and \(s_i^e\): \(s_o^e = 12.0 - 7.5 = 4.5 \mathrm{cm}\) So, Jordan should place the lenses \(4.5 \mathrm{cm}\) apart.
06

Calculate the magnification for the eyepiece lens

Since the object distance for the eyepiece lens is placed at the focal point, the angular magnification of the eyepiece lens is happening at infinity. Thus, we can use the following formula for the angular magnification of the eyepiece lens: \(M_e = 1 + \frac{f_e}{D}\), where \(D\) is the nearest clear vision distance, which is typically assumed to be \(25 \mathrm{cm}\). Now, plug in the value of \(f_e\): \(M_e = 1 + \frac{7.5}{25} = 1.3\)
07

Calculate the total angular magnification of the microscope

The total angular magnification of the microscope is the product of the magnification of the objective lens and the angular magnification of the eyepiece lens: \(M_{total} = M_o \times M_e\) Now, plug in the values of \(M_o\) and \(M_e\): \(M_{total} = 7.5 \times 1.3 \approx 9.75\) The angular magnification of the microscope will be approximately \(9.75\) times.

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Most popular questions from this chapter

The Ortiz family is viewing slides from their summer vacation trip to the Grand Canyon. Their slide projector has a projection lens of \(10.0-\mathrm{cm}\) focal length and the screen is located \(2.5 \mathrm{m}\) from the projector. (a) What is the distance between the slide and the projection lens? (b) What is the magnification of the image? (c) How wide is the image of a slide of width \(36 \mathrm{mm}\) on the screen? (See the figure with Problem 16 .)
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