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Chapter 34: Problem 50

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

Understanding the Problem

The problem involves viewing an insect using a magnifier. The insect is positioned at the focal point of the magnifier, and we need to determine the focal length that will give the insect an angular size of 0.025 radians.
02

Identifying the Relationship

When an object is placed at the focal point of a magnifier, the angular size or magnification can be calculated using the formula for angular magnification: \( M = \frac{h}{f} \), where \( h \) is the size of the object (in meters) and \( f \) is the focal length. The angular size \( \theta \) can be expressed as \( \theta = \frac{h}{f} \).
03

Converting Units

The length of the insect is given as 2.00 mm. Convert this length to meters because SI units are required for the calculation. So, \( 2.00 \text{ mm} = 0.002 \text{ m} \).
04

Using the Equation

Substitute the values into the equation \( \theta = \frac{h}{f} \). We know \( \theta = 0.025 \) radians and \( h = 0.002 \text{ m} \). The equation becomes \( 0.025 = \frac{0.002}{f} \).
05

Solving for Focal Length

Rearrange the equation \( 0.025 = \frac{0.002}{f} \) to solve for \( f \). Multiply both sides by \( f \) to get \( 0.025f = 0.002 \). Then divide both sides by 0.025 to get \( f = \frac{0.002}{0.025} \).
06

Final Calculation

Calculate \( f = \frac{0.002}{0.025} = 0.08 \). Thus, the focal length needed is 0.08 meters (or 80 mm).

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Key Concepts

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

Angular Magnification
Angular magnification is a key concept in optics, especially when using magnifying glasses or lenses. It refers to how much larger an object appears when viewed through a lens compared to viewing it with the naked eye.
The magnification formula for lenses is given by:
  • The formula is: \( M = \frac{\theta}{\theta_0} \)
  • \( M \) is the angular magnification.
  • \( \theta \) is the angular size of the image.
  • \( \theta_0 \) is the angular size of the object when viewed without magnification.
When an object is placed at the focal point of the lens, the magnification can also be expressed as:\[ M = \frac{h}{f} \]where \( h \) is the height (or size) of the object and \( f \) is the focal length of the lens.
Understanding angular magnification helps in selecting the right lens to achieve the desired level of detail for small objects like insects.
Focal Length
The focal length of a lens is the distance from the lens to the point where light rays converge to form a sharp image. It's a fundamental property of lenses and determines how strongly the lens can magnify an object.
For example, if you're observing an insect using a magnifier, you need to place the insect at the lens's focal point. This setup allows the lens to enlarge the image effectively.
The calculation for focal length in optical instruments involves the relationship:
  • When the angular size \( \theta \) of the object is known along with the size \( h \) of the object, we can use the relationship: \( \theta = \frac{h}{f} \).
  • Solving this for \( f \), we get: \( f = \frac{h}{\theta} \).
Selecting an appropriate focal length is crucial to achieving the desired magnification and clarity of the image.
Angular Size
Angular size is an important concept that describes how large an object appears from a certain point of view. It is measured in radians and can vary depending on the distance and the actual size of the object.
In optics, when viewing an object through a lens, the angular size allows us to understand how much of the object is visible through the lens.
The formula for angular size is:
  • \( \theta = \frac{h}{d} \)
  • \( \theta \) is the angular size in radians.
  • \( h \) is the height or size of the object.
  • \( d \) is the distance from the observer to the object.
When dealing with small objects, knowing their angular size assists in determining the needed focal length to view them with the desired magnification.
Convert Units to SI
The International System of Units (SI) is the standard system of measurement used worldwide. When dealing with scientific calculations, it's essential to convert any given units to SI units to ensure accuracy and consistency.
In optics, lengths are usually converted to meters for calculations.
  • For example, if an object's length is 2.00 mm, it should be converted to meters as follows: 2.00 mm = 0.002 m.
  • This conversion is done by recognizing that 1 mm is equal to 0.001 meters.
Applying SI units simplifies equations and makes it easier to derive accurate results. In optical exercises, converting measurements to SI is a crucial step that precedes any calculations.

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Most popular questions from this chapter

Spherical aberration is a blurring of the image formed by a spherical mirror. It occurs because parallel rays striking the mirror far from the optic axis are focused at a different point than are rays near the axis. This problem is usually minimized by using only the center of a spherical mirror. (a) Show that for a spherical concave mirror, the focus moves toward the mirror as the parallel rays move toward the outer edge of the mirror. (Hint: Derive an analytic expression for the distance from the vertex to the focus of the ray for a particular parallel ray. This expression should be in terms of (i) the radius of curvature \(R\) of the mirror and (ii) the angle \(\boldsymbol{\theta}\) between the incident ray and a line connecting the center of curvature of the mirror with the point where the ray strikes the mirror (b) What value of \(\theta\) produces a 2\(\%\) change in the location of the focus compared to the location for \(\theta\) very close to zero?

Choosing a Camera Lens. The picture size on ordinary \(35-\mathrm{mm}\) camera film is \(24 \mathrm{mm} \times 36 \mathrm{mm}\) . Focal lengths of lenses available for \(35-\mathrm{mm}\) cameras typically include \(28,35,50\) (the "normal" lens), \(85,100,135,200,\) and \(300 \mathrm{mm},\) among others. Which of these lenses should be used to photograph the following objects, assuming that the object is to fill most of the picture area? (a) a building 240 \(\mathrm{m}\) tall and 160 \(\mathrm{m}\) wide at a distance of 600 \(\mathrm{m}\) , and (b) a mobile home 9.6 \(\mathrm{m}\) in length at a distance of 40.0 \(\mathrm{m}\) .

A microscope is focused on the upper surface of a glass plate. A second plate is then placed over the first. To focus on the bottom surface of the second plate, the microscope must be raised 0.780 \(\mathrm{mm}\) . To focus on the upper surface, it must be raised another 2.50 \(\mathrm{mm}\) . Find the index of refraction of the second plate.

The left end of a long glass rod 6.00 \(\mathrm{cm}\) in diameter has a convex hemispherical surface 3.00 \(\mathrm{cm}\) in radius. The refractive index of the glass is 1.60 . Determine the position of the image if an object is placed in air on the axis of the rod at the following distances to the left of the vertex of the curved end: (a) infinitely far, (b) \(12.0 \mathrm{cm} ;(\mathrm{c}) 2.00 \mathrm{cm} .\)

Three-Dimensional Image. The longitudinal magnification is defined as \(m^{\prime}=d s^{\prime} / d s .\) It relates the longitudinal dimension of a small object to the longitudinal dimension of its image. (a) Show that for a spherical mirror, \(m^{\prime}=-m^{2}\) . What is the significance of the fact that \(m^{\prime}\) is always negative? (b) A wire frame in the form of a small cube 1.00 \(\mathrm{mm}\) on a side is placed with its center on the axis of a concave mirror with radius of curvature 150.0 \(\mathrm{cm}\) . The sides of the cube are all either parallel or perpendicular to the axis. The cube face toward the mirror is 200.0 \(\mathrm{cm}\) to the left of the mirror vertex. Find (i) the location of the image of this face and of the opposite face of the cube; (ii) the lateral and longitudinal magnifications; (iii) the shape and dimensions of each of the six faces of the image.
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