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You want to view through a magnifier an insect that is 2.00 mm long. 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 0.032 radian?

Short Answer

Expert verified
The focal length is 62.5 mm.

Step by step solution

01

Identify known values

We are given the insect length of 2.00 mm and the angular size of the image as 0.032 radian. We need to find the focal length of the magnifier.
02

Relate angular size to image size

The angular size \( \theta \) of an object at the focal point of a lens is given by \( \theta = \frac{h}{f} \), where \( h \) is the size of the object (2.00 mm) and \( f \) is the focal length (unknown).
03

Solve for focal length

Rearrange the formula from step 2 to solve for the focal length: \( f = \frac{h}{\theta} = \frac{2.00\, \text{mm}}{0.032\, \text{radian}} \).
04

Calculate the value

Perform the division to find \( f \):\[ f = \frac{2.00}{0.032} = 62.5 \text{ mm} \].
05

Verify Units and Solution

Ensure the calculated focal length makes sense physically. A focal length of 62.5 mm gives the desired angular size of 0.032 radians for the particular object size of 2.00 mm.

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Key Concepts

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

Focal Length
The concept of focal length is essential in geometric optics, as it defines how lenses focus light. The focal length is the distance from the lens to the focal point, where parallel rays of light meet after passing through the lens. In the context of a magnifier, the focal length determines how much an object will appear enlarged or reduced when viewed through the lens.
In our specific exercise, we have a lens used as a magnifier to view an insect. We need a particular focal length to achieve the given angular size for the object when viewed through the magnifier. By positioning the insect at the lens's focal point, the focal length ensures that the light rays converge correctly to give an optimal magnified view.
  • An appropriate focal length enables clear and precise imaging.
  • The object appears at the focal point, creating ideal magnification.
Understanding focal length helps in various applications, including photography, eyewear correction, and, as seen here, microscopy.
Angular Magnification
Angular magnification is the measure of how much larger an object appears when seen through a lens compared to when viewed with the naked eye. It’s crucial in understanding how lenses like magnifiers enhance the apparent size of objects. This concept is especially applicable in our problem scenario, where a magnifier is used to view an insect.
Angular magnification describes the ratio of the angular size of the image to the angular size of the object without the magnifier. In our case, the goal is to create an image with an angular size of 0.032 radians. This is achieved through the suitable adjustment of the focal length, enabling the object to fill more of the observer's field of view.
  • Angular magnification depends directly on the focal length and the object's position relative to the lens.
  • A higher magnification results in a larger image and greater detail visibility.
This principle is widely used in optical tools like telescopes and microscopes for detailed observations.
Lens Formula
The lens formula is a fundamental equation in geometric optics that relates focal length, object distance, and image distance. The lens formula is expressed as \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length, \( d_o \) is the object distance from the lens, and \( d_i \) is the image distance from the lens.
This formula helps us understand how images are formed by the lens. By knowing any two of the parameters, we can calculate the third. However, for our specific case, where the object is at the focal point, the formula simplifies as the image is formed at infinity, providing maximum angular magnification.
  • The lens formula allows calculation flexibility for various optical scenarios.
  • It's crucial for designing and understanding lenses in devices like cameras and glasses.
By mastering the lens formula, we gain insights into the behavior of lenses across numerous real-world applications.

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Most popular questions from this chapter

A person swimming 0.80 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 m above the swimmer. What is the actual height of the diving board above the surface of the water?

Contact lenses are placed right on the eyeball, so the distance from the eye to an object (or image) is the same as the distance from the lens to that object (or image). A certain person can see distant objects well, but his near point is 45.0 cm from his eyes instead of the usual 25.0 cm. (a) Is this person nearsighted or farsighted? (b) What type of lens (converging or diverging) is needed to correct his vision? (c) If the correcting lenses will be contact lenses, what focal length lens is needed and what is its power in diopters?

A converging lens with a focal length of 70.0 cm forms an image of a 3.20-cm- tall real object that is to the left of the lens. The image is 4.50 cm tall and inverted. Where are the object and image located in relation to the lens? Is the image real or virtual?

A speck of dirt is embedded 3.50 cm below the surface of a sheet of ice \((n = 1.3092)\). What is its apparent depth when viewed at normal incidence?

You hold a spherical salad bowl 60 cm in front of your face with the bottom of the bowl facing you. The bowl is made of polished metal with a 35 cm radius of curvature. (a) Where is the image of your 5.0 cm tall nose located? (b) What are the image's size, orientation, and nature (real or virtual)?

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