/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 60 You want to view an insect 2.00 ... [FREE SOLUTION] | 91Ó°ÊÓ

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You want to view an insect 2.00 \(\mathrm{mm}\) in length through a magnifier. If the insect is to be at the focal point of the magnifier, what focal length will give the image of the insect an angular size of 0.025 radian?

Short Answer

Expert verified
The focal length required is 80 mm.

Step by step solution

01

Understand the problem

We need to find the focal length of a magnifier such that an insect with a length of 2.00 mm appears with an angular size of 0.025 radians when it is placed at the focal point of the magnifier.
02

Use the angular magnification formula

The angular magnification, when an object is at the focal point, is given by \[ M = \frac{\theta}{\theta_0} = \frac{s_i}{s_o} \] where \( \theta \) is the angular size of the image, \( \theta_0 \) is the angular size of the object with the naked eye, \( s_i \) is the image size, and \( s_o \) is the object size.
03

Relate image size to focal length

For an object placed at the focal point of a lens, the angular size of the image \( \theta \) is given by \[ \theta = \frac{s}{f} \] where \( s \) is the size of the object and \( f \) is the focal length. Here, \( s = 2.00 \ mm \) and \( \theta = 0.025 \ rad \).
04

Solve for the focal length

We can rearrange the equation \( \theta = \frac{s}{f} \) to solve for \( f \): \[ f = \frac{s}{\theta} \] Substitute \( s = 2.00 \ mm \) and \( \theta = 0.025 \ rad \) into the equation: \[ f = \frac{2.00 \ mm}{0.025} = 80 \ mm \]
05

Verify the result

Check the calculation to ensure that having a focal length of 80 mm will indeed produce the desired angular size when the insect is at the focal point. The calculations are consistent with the expectations from the lens equation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angular Magnification
Angular magnification plays a central role in how we perceive the size of objects through a lens. Imagine you are looking at a tiny insect through a magnifier. Without the magnification, you'd see the insect and its details in its actual size, which might be too small to appreciate fully. But a magnifier essentially makes the insect appear larger by increasing its angular size.The magnification factor is defined by the ratio of the angular size of the image to the angular size of the object as seen with the naked eye. This relationship is expressed mathematically by the formula:- \( M = \frac{\theta}{\theta_0} \) - where \( \theta \) is the angular size through the lens and \( \theta_0 \) is the angular size without the lens.Understanding this simple mathematical relation helps us predict how large an object will appear when viewed through different lenses, which is invaluable for designing and using optical tools effectively.
Focal Length Calculation
Calculating the focal length of a lens is crucial because it determines how strongly the lens can magnify objects. If you want to magnify an object, like an insect, to a certain angular size, you need to adjust the lens's focal length accordingly.The relationship between the focal length \( f \), the size of the object \( s \), and the angular size of the image \( \theta \) is critical. The formula that links these variables is:- \( \theta = \frac{s}{f} \)In this equation, \( \theta \) represents the desired angular size, \( s \) is the actual size of the object, and \( f \) is the focal length. To find \( f \), you simply rearrange this equation to:- \( f = \frac{s}{\theta} \)By substituting the known values for \( s \) and \( \theta \) into this formula, you can find the focal length required for a specific magnification. For the insect example, with \( s = 2.00 \ mm \) and \( \theta = 0.025 \ rad \), the focal length comes out to 80 mm, meaning this lens will perfectly magnify the insect to the desired scale.
Image Formation with Lenses
The process of image formation with lenses involves the bending of light rays toward or away from a principal focus, based on the lens type. In our scenario, the goal is to make the insect appear larger through a magnifier. When we place the object at the focal point of a lens: - The light rays exiting the lens are parallel. - This arrangement gives rise to an image that seems to come from a very large or infinite distance, making it appear significantly larger than the actual object. The image created by the lens does not exist on just the opposite side like a real image, but it's perceived directly by the observer's eyes. This concept is essential when discussing magnifiers, telescopes, or any optical device that intends to enlarge the perceived size of an object. Each lens has particular properties that influence how an image is formed, such as the shape of the lens and its material. By understanding these properties, we can better appreciate how lenses are designed to optimize the clarity and size of the images they produce.

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Most popular questions from this chapter

A lens forms an image of an object. The object is 16.0 \(\mathrm{cm}\) from the lens. The image is 12.0 \(\mathrm{cm}\) from the lens on the same side as the object. (a) What is the focal length of the lens? Is the lens converging or diverging? (b) If the object is 8.50 \(\mathrm{mm}\) tall, how tall is the image? Is it erect or inverted? (c) Draw a principal-ray diagram.

An object is 16.0 \(\mathrm{cm}\) to the left of a lens. The lens forms an image 36.0 \(\mathrm{cm}\) to the right of the lens. (a) What is the focal length of the lens? Is the lens converging or diverging? (b) If the object is 8.00 mm tall, how tall is the image? Is it erect or inverted? (c) Draw a principal-ray diagram.

A person swimming 0.80 \(\mathrm{m}\) below the surface of the water in a swimming pool looks at the diving board that is directly overhead and sees the image of the board that is formed by refraction at the surface of the water. This image is a height of 5.20 \(\mathrm{m}\) above the swimmer. What is the actual height of the diving board above the surface of the water?

An object is placed 18.0 \(\mathrm{cm}\) from a screen. (a) At what two points between object and screen may a converging lens with a 3.00 -cm focal length be placed to obtain an image on the screen? (b) What is the magnification of the image for each position of the lens?

A converging lens with a focal length of 7.00 \(\mathrm{cm}\) forms an image of a \(4.00-\mathrm{mm}\)-tall real object that is to the left of the lens. The image is 1.30 \(\mathrm{cm}\) tall and erect. Where are the object and image located? Is the image real or virtual?

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