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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

Two lenses, separated by a distance of \(21.0 \mathrm{cm},\) are used in combination. The first lens has a focal length of \(+30.0 \mathrm{cm} ;\) the second has a focal length of \(-15.0 \mathrm{cm} .\) An object, $2.0 \mathrm{mm}\( long, is placed \)1.8 \mathrm{cm}$ before the first lens. (a) What are the intermediate and final image distances relative to the corresponding lenses? (b) What is the total magnification? (c) What is the height of the final image?
The area occupied by one frame on 35 -mm film is \(24 \mathrm{mm}\) by $36 \mathrm{mm}-\( see the figure with Problem \)16 .$ The focal length of the camera lens is \(50.0 \mathrm{mm}\). A picture is taken of a person $182 \mathrm{cm}$ tall. What is the minimum distance from the camera for the person to stand so that the image fits on the film? Give two answers; one for each orientation of the camera.
(a) What is the angular size of the Moon as viewed from Earth's surface? See the inside back cover for necessary information. (b) The objective and eyepiece of a refracting telescope have focal lengths \(80 \mathrm{cm}\) and \(2.0 \mathrm{cm}\) respectively. What is the angular size of the Moon as viewed through this telescope?
You plan to project an inverted image \(30.0 \mathrm{cm}\) to the right of an object. You have a diverging lens with focal length \(-4.00 \mathrm{cm}\) located \(6.00 \mathrm{cm}\) to the right of the object. Once you put a second lens at \(18.0 \mathrm{cm}\) to the right of the object, you obtain an image in the proper location. (a) What is the focal length of the second lens? (b) Is this lens converging or diverging? (c) What is the total magnification? (d) If the object is \(12.0 \mathrm{cm}\) high. what is the image height?
Unless the problem states otherwise, assume that the distance from the comea- lens system to the retina is \(2.0 \mathrm{cm}\) and the normal near point is \(25 \mathrm{cm}.\) If the distance from the lens system (cornea + lens) to the retina is $2.00 \mathrm{cm},$ show that the focal length of the lens system must vary between \(1.85 \mathrm{cm}\) and \(2.00 \mathrm{cm}\) to see objects from $25.0 \mathrm{cm}$ to infinity.
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