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Keesha is looking at a beetle with a magnifying glass. She wants to see an upright, enlarged image at a distance of \(25 \mathrm{cm} .\) The focal length of the magnifying glass is \(+5.0 \mathrm{cm} .\) Assume that Keesha's eye is close to the magnifying glass. (a) What should be the distance between the magnifying glass and the beetle? (b) What is the angular magnification? (tutorial: magnifying glass II).

Short Answer

Expert verified
Answer: The distance between the magnifying glass and the beetle is 6.25 cm, and the angular magnification of the magnifying glass is 5.

Step by step solution

01

Recall the lens formula

First, we need to recall the lens formula, which relates the object distance (p), the image distance (q), and the focal length (f) as follows: \( \frac{1}{f} = \frac{1}{p} + \frac{1}{q} \)
02

Plug in the values

We are given that the image distance (q) is \(25 \mathrm{cm}\) and the focal length (f) is \(+5.0 \mathrm{cm}\). Plugging these values into the lens formula, we get: \( \frac{1}{+5.0} = \frac{1}{p} + \frac{1}{25} \)
03

Solve for the object distance

To find the object distance (p), we can rearrange and solve the equation: \( \frac{1}{p} = \frac{1}{+5.0} - \frac{1}{25} \) \( \frac{1}{p} = \frac{4}{25} \) \( p = \frac{25}{4} \) \( p = 6.25 \mathrm{cm} \) So the distance between the magnifying glass and the beetle should be \(6.25 \mathrm{cm}\).
04

Recall the formula for angular magnification

Next, let's find the angular magnification. Recall the formula for angular magnification (M): \( M = \frac{25 \mathrm{cm}}{f} \)
05

Plug in the values and solve for angular magnification

Now, we can plug in the values for the image distance (q) and the focal length (f) to find the angular magnification: \( M = \frac{25}{5.0} \) \( M = 5 \) The angular magnification of the magnifying glass is 5. So, the distance between the magnifying glass and the beetle should be \(6.25 \mathrm{cm}\), and the angular magnification is 5.

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Most popular questions from this chapter

An object is located \(16.0 \mathrm{cm}\) in front of a converging lens with focal length \(12.0 \mathrm{cm} .\) To the right of the converging lens, separated by a distance of \(20.0 \mathrm{cm},\) is a diverging lens of focal length \(-10.0 \mathrm{cm} .\) Find the location of the final image by ray tracing and verify using the lens equations.
Two converging lenses, separated by a distance of \(50.0 \mathrm{cm},\) are used in combination. The first lens, located to the left, has a focal length of \(15.0 \mathrm{cm} .\) The second lens, located to the right, has a focal length of \(12.0 \mathrm{cm} .\) An object, \(3.00 \mathrm{cm}\) high, is placed at a distance of \(20.0 \mathrm{cm}\) in front of the first lens. (a) Find 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?
You would like to project an upright image at a position \(32.0 \mathrm{cm}\) to the right of an object. You have a converging lens with focal length $3.70 \mathrm{cm}\( located \)6.00 \mathrm{cm}$ to the right of the object. By placing a second lens at \(24.65 \mathrm{cm}\) 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?
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