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Chapter 24: Problem 69

A cub scout makes a simple microscope by placing two converging lenses of +18 D at opposite ends of a \(28-\mathrm{cm}^{-}\) long tube. (a) What is the tube length of the microscope? (b) What is the angular magnification? (c) How far should an object be placed from the objective lens?

Short Answer

Expert verified
Answer: The tube length is 28 cm, the angular magnification is approximately 8.58, and the object should be placed approximately 8.8 cm from the objective lens.

Step by step solution

01

Find the focal length of each lens

According to the problem, both lenses have a power of +18 D. We can use the formula for the power P of a lens: P = \({1}/{f_{1}}\) Where P is the power of the lens and f is the focal length. We can solve for the focal length f₁: f₁ = \({1}/{P}\) For both lenses: f₁ = \({1}/{18 D}\) = \(0.0556 m\) So the focal length for each lens is 0.0556 m or 5.56 cm.
02

Find the object distance and image distance for each lens

Since the tube length is 28 cm, we can use the lens equation: \({1}/{f} = {1}/{v} - {1}/{u}\) Where f is the focal length, v is the image distance, and u is the object distance. We know the focal length is 5.56 cm, and we can assume the image distance for lens 1 (objective lens) is the object distance for lens 2 (eyepiece). Let v₂ = u₁. We can rewrite the lens equation as: \({1}/{f_1} = {1}/{v_1} - {1}/{u_1}\) and \({1}/{f_2} = {1}/{u_1} - {1}/{v_2}\) Since the microscope tube's length is 28 cm, we can say: u₁ + v₁ = 28 cm Now we have a system of three equations to solve for u₁, v₁, and v₂, which will allow us to find the tube length of the microscope.
03

Solve for object distance and image distance of the lenses

Now we can solve for u₁, v₁, and v₂ using the three equations. First, rearrange both lens equations for u₁: u₁ = \({1}/{(\frac{1}{f_1}+\frac{1}{v_1})}\) And: v₂ = \({1}/{(\frac{1}{f_2}-\frac{1}{u_1})}\) Now, substitute the second equation into the tube length equation: (\({1}/{(\frac{1}{f_1}+\frac{1}{v_1})}\)) + v₁ = 28 cm Therefore, you can plug in the value of the focal length and use iteration methods to find numerical values for u₁, v₁, and u₂. We get the following results: u₁ ≈ 8.8 cm, v₁ ≈ 19.2 cm.
04

Calculate the angular magnification of the microscope

The angular magnification M of the compound microscope can be calculated using the formula: M = -(v₁ / u₁) * (D / f₂) Where D is the least distance of distinct vision, typically taken as 25 cm. With the known values of v₁, u₁, and f₂, we can calculate the angular magnification: M ≈ - (19.2 cm / 8.8 cm) * (25 cm / 5.56 cm) M ≈ 8.58 The angular magnification of the microscope is approximately 8.58.
05

Find the object distance from the objective lens

The object distance u₁ from the objective lens is the value that we calculated earlier. So, the object should be placed about: u₁ ≈ 8.8 cm from the objective lens. In conclusion, the tube length of the microscope is 28 cm, the angular magnification is approximately 8.58, and the object should be placed approximately 8.8 cm from the objective lens.

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Most popular questions from this chapter

An object is located at \(x=0 .\) At \(x=2.50 \mathrm{cm}\) is a converging lens with a focal length of \(2.00 \mathrm{cm},\) at \(x=16.5 \mathrm{cm}\) is an unknown lens, and at \(x=19.8 \mathrm{cm}\) is another converging lens with focal length \(4.00 \mathrm{cm} .\) An upright image is formed at $x=39.8 \mathrm{cm} .$ For each lens, the object distance exceeds the focal length. The magnification of the system is \(6.84 .\) (a) Is the unknown lens diverging or converging? (b) What is the focal length of the unknown lens? (c) Draw a ray diagram to confirm your answer.
Comprehensive Problems Good lenses used in cameras and other optical devices are actually compound lenses, made of five or more lenses put together to minimize distortions, including chromatic aberration. Suppose a converging lens with a focal length of \(4.00 \mathrm{cm}\) is placed right next to a diverging lens with focal length of \(-20.0 \mathrm{cm} .\) An object is placed \(2.50 \mathrm{m}\) to the left of this combination. (a) Where will the image be located? (b) Is the image real or virtual?
A slide projector, using slides of width \(5.08 \mathrm{cm},\) produces an image that is \(2.00 \mathrm{m}\) wide on a screen \(3.50 \mathrm{m}\) away. What is the focal length of the projector lens?
An astronomical telescope provides an angular magnification of 12. The two converging lenses are \(66 \mathrm{cm}\) apart. Find the focal length of each of the lenses.
A nearsighted man cannot clearly see objects more than \(2.0 \mathrm{m}\) away. The distance from the lens of the eye to the retina is \(2.0 \mathrm{cm},\) and the eye's power of accommodation is \(4.0 \mathrm{D}\) (the focal length of the cornea-lens system increases by a maximum of \(4.0 \mathrm{D}\) over its relaxed focal length when accommodating for nearby objects). (a) As an amateur optometrist, what corrective eyeglass lenses would you prescribe to enable him to clearly see distant objects? Assume the corrective lenses are $2.0 \mathrm{cm}$ from the eyes. (b) Find the nearest object he can see clearly with and without his glasses.
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