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Chapter 34: Problem 60

You want to view through a magnifier an insect that is \(2.00 \mathrm{~mm}\) 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 needed for the magnifier is 15.625 mm.

Step by step solution

01

Understand the question and interpret the values

First, interpret the values given in the question. The length of the insect is 2.00mm (or 0.002m) and the magnified image size is 0.032 rad. The near point distance (distance from the eye at which an object can be clearly focused) is conventionally taken as 25cm or 0.25m.
02

Apply the formula for angular magnification

In this step, we apply the formula for angular magnification. Angular magnification (m) is equal to the object size (h) divided by its distance from the lens (f), then multiplied with the distance of near point (D), i.e., m = \( \frac{D \cdot h}{f} \). We rearrange to make f the subject: f = \( \frac{D \cdot h}{m} \).
03

Substituting the given values

Substitute the values into the formula: f = \( \frac{0.25 \cdot 0.002}{0.032} \).
04

Calculating the focal length

By performing the calculation, you should find that f = 0.015625 meters or 15.625 mm.

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Key Concepts

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

Focal Length
The focal length of a lens is a crucial concept in optics, as it defines how strongly a lens converges or diverges light. When using a magnifying lens to view an object, such as an insect, one considers the focal length to determine how large the image will appear.

The focal length (\( f \)) is the distance between the lens and its focal point, which is the point where parallel rays of light either converge or appear to diverge after passing through the lens. A shorter focal length means a stronger lens that bends light more sharply, creating a larger magnified image when the object is placed at this point. In the exercise, we found the focal length needed by rearranging the angular magnification formula to solve for \( f \).

In practical terms:
  • A shorter focal length provides higher magnification.
  • The focal length is measured in meters or millimeters.
  • It determines the size and clarity of the image produced.
Optics
Optics is the branch of physics that studies light and its interactions with various materials, including lenses and mirrors. It encompasses concepts like refraction, reflection, and the behavior of light as it passes through different media. Understanding optics is essential for solving problems related to lens magnification, as these concepts dictate how images are formed.

In our exercise, the use of a lens to magnify an insect relies on fundamental optical principles. When placed at a lens's focal point, the light from the object does not converge or diverge but instead exits parallel on the other side, creating a virtual image at infinity, which appears larger to the viewer.

Key points in optics related to the exercise:
  • Light changes direction when passing through lenses, known as refraction.
  • Lenses have two main types: converging (convex) and diverging (concave).
  • The focal length and lens curvature influence the amount of magnification.
Lens Formula
The lens formula provides a relationship between the object distance (\( u \)), image distance (\( v \)), and the focal length (\( f \)) of a lens. It is expressed as:\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]This formula allows us to calculate any one of these three parameters if the other two are known, making it an essential tool in understanding how lenses work.

In the context of angular magnification, we focused primarily on the focal length. However, understanding the broader lens formula aids in comprehending how changing focal lengths will alter image properties.

Fundamental points about the lens formula include:
  • The lens formula is applicable for thin lenses and is essential for optical calculations.
  • It helps in predicting the location and size of an image produced by a lens.
  • Integration with the magnification concept provides comprehensive understanding.

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Most popular questions from this chapter

An insect \(3.75 \mathrm{~mm}\) tall is placed \(22.5 \mathrm{~cm}\) to the left of a thin planoconvex lens. The left surface of this lens is flat, the right surface has a radius of curvature of magnitude \(13.0 \mathrm{~cm},\) and the index of refraction of the lens material is \(1.70 .\) (a) Calculate the location and size of the image this lens forms of the insect. Is it real or virtual? Erect or inverted? (b) Repeat part (a) if the lens is reversed.

A candle \(4.85 \mathrm{~cm}\) tall is \(39.2 \mathrm{~cm}\) to the left of a plane mirror. Where is the image formed by the mirror, and what is the height of this image?

When an object is placed at the proper distance to the left of a converging lens, the image is focused on a screen \(30.0 \mathrm{~cm}\) to the right of the lens. A diverging lens is now placed \(15.0 \mathrm{~cm}\) to the right of the converging lens, and it is found that the screen must be moved \(19.2 \mathrm{~cm}\) farther to the right to obtain a sharp image. What is the focal length of the diverging lens?

A speck of dirt is embedded \(3.50 \mathrm{~cm}\) below the surface of a sheet of ice \((n=1.309) .\) What is its apparent depth when viewed at normal incidence?

A pinhole camera is just a rectangular box with a tiny hole in one face. The film is on the face opposite this hole, and that is where the image is formed. The camera forms an image without a lens. (a) Make a clear ray diagram to show how a pinhole camera can form an image on the film without using a lens. (Hint: Put an object outside the hole, and then draw rays passing through the hole to the opposite side of the box.) (b) A certain pinhole camera is a box that is \(25 \mathrm{~cm}\) square and \(20.0 \mathrm{~cm}\) deep, with the hole in the middle of one of the \(25 \mathrm{~cm} \times 25 \mathrm{~cm}\) faces. If this camera is used to photograph a fierce chicken that is \(18 \mathrm{~cm}\) high and \(1.5 \mathrm{~m}\) in front of the camera, how large is the image of this bird on the film? What is the lateral magnification of this camera?
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