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Chapter 25: Problem 42

You are examining a flea with a converging lens that has a focal length of \(4.00 \mathrm{~cm}\). If the image of the flea is 6.50 times the size of the flea, how far is the flea from the lens? Where, relative to the lens, is the image?

Short Answer

Expert verified
The flea is 3.85 cm from the lens, and the image is 25.03 cm behind the lens.

Step by step solution

01

Identifying Given Values and Formulas

We have a converging lens with a focal length of \( f = 4.00 \text{ cm} \) and an image magnification of 6.5 times, which means the magnification \( m = 6.5 \). The formula for magnification is \( m = \frac{h'}{h} = \frac{-q}{p} \), where \( h' \) and \( h \) are the image and object heights respectively, and \( q \) and \( p \) are the image and object distances. Also, the lens formula is \( \frac{1}{f} = \frac{1}{p} + \frac{1}{q} \).
02

Using the Magnification Formula

The formula for magnification \( m = \frac{-q}{p} = 6.5 \) gives us \( q = -6.5p \) because the image is inverted and real. This equation links image distance \( q \) to object distance \( p \).
03

Substitute into Lens Formula

Substituting \( q = -6.5p \) into the lens formula \( \frac{1}{f} = \frac{1}{p} + \frac{1}{q} \), we have \( \frac{1}{4} = \frac{1}{p} + \frac{1}{-6.5p} \).
04

Solve for Object Distance \( p \)

Simplify the equation to get a common denominator: \( \frac{1}{4} = \frac{-6.5 + 1}{-6.5p} \). This becomes \( \frac{1}{4} = \frac{-5.5}{-6.5p} \). Cross-multiplied gives: \( 4 \times -6.5p = -5.5 \), hence \( p = \frac{5.5}{26} \approx 3.85 \text{ cm} \).
05

Calculate the Image Distance \( q \)

To find where the image is, substitute \( p = 3.85 \) cm back into \( q = -6.5p \): \( q = -6.5 \times 3.85 \approx -25.03 \text{ cm} \).
06

Determine Relative Position

Since \( q \) is negative, the image is on the same side as the object, which is a characteristic of real images formed by a converging lens when the object is outside the focal length.

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Key Concepts

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

Converging Lens
A converging lens, also known as a convex lens, is a lens that bends light rays towards each other as they pass through it. These lenses are thicker in the middle than at the edges. When parallel light rays enter a converging lens, they are refracted towards the principal axis and eventually meet at a point called the focal point.

Converging lenses are commonly used in various optical devices such as microscopes, cameras, and eyeglasses. They help focus light to produce clear images, which can be larger or smaller than the object itself. This ability makes them perfect for magnifying objects and playing a crucial role in many optical instruments.
  • Key Characteristics: Bends light rays towards each other
  • Applications: Used in cameras, projectors, and corrective lenses
  • Functionality: Form real or virtual images depending on the object's position
Focal Length
The focal length of a lens is a measure of how strongly it converges or diverges light. It's the distance from the lens to the focal point, where light rays converge. For a converging lens, the focal length is positive, which is crucial in determining the nature of the image formed.

In our exercise, the given focal length is 4.00 cm. This indicates that any parallel light rays entering the lens will converge at a point 4.00 cm from the lens. The focal length is a vital factor in deciding how the lens will manipulate light, affecting the size and distance of the image that forms.
  • Definition: Positive for converging lenses
  • Importance: Determines the image properties
  • Given Example: Focal length of 4.00 cm results in specific image characteristics
Magnification
Magnification refers to the ratio of the image size to the object size. It's a dimensionless number that tells us how much larger or smaller the image is compared to the actual object. In optics, negative magnification can indicate an inverted image, which is typical for real images formed with a converging lens.

In the given problem, the magnification is 6.5 times. This means the image is 6.5 times larger than the object itself. Such a high magnification signifies that the object is close to the lens and placed at a specific location relative to the focal point.
  • Understanding: Ratio of image size to object size
  • Negative Sign: Indicates image inversion
  • Example Case: 6.5 magnification suggests a large, real, and inverted image
Lens Formula
The lens formula is a vital equation in optics that connects the focal length (\( f \)), the object distance (\( p \)), and the image distance (\( q \)). It is represented as \( \frac{1}{f} = \frac{1}{p} + \frac{1}{q} \). This formula allows us to calculate any one of these three variables if the other two are known.

In practice, the lens formula is used to determine where the image will form and how large it will be. By substituting known values into this formula, you can solve for unknown distances, helping you visualize how the lens alters the path of light to produce images.
  • Equation: \( \frac{1}{f} = \frac{1}{p} + \frac{1}{q} \)
  • Purpose: Connects object distance, image distance, and focal length
  • Applications: Essential for solving optical problems and understanding image formation

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Most popular questions from this chapter

(a) A small refracting telescope designed for individual use has an objective lens with a diameter of \(6.00 \mathrm{~cm}\) and a focal length of \(1.325 \mathrm{~m}\). What is the \(f\) -number of this instrument? (b) The 200 -inchdiameter objective mirror of the Mount Palomar telescope has an \(f\) -number of \(3.3 .\) Calculate its focal length. (c) The distance between lens and retina for a normal human eye is about \(2.50 \mathrm{~cm},\) and the pupil can vary in size from \(2.0 \mathrm{~mm}\) to \(8.0 \mathrm{~mm} .\) What is the range of \(f\) -numbers for the human eye?

A person with a near point of \(85 \mathrm{~cm},\) but excellent distant vision, normally wears corrective glasses. But he loses them while traveling. Fortunately, he has his old pair as a spare. (a) If the lenses of the old pair have a power of +2.25 diopters, what is his near point (measured from his eye) when he is wearing the old glasses if they rest \(2.0 \mathrm{~cm}\) in front of his eye? (b) What would his near point be if his old glasses were contact lenses instead?

A certain digital camera having a lens with focal length \(7.50 \mathrm{~cm}\) focuses on an object \(1.85 \mathrm{~m}\) tall that is \(4.25 \mathrm{~m}\) from the lens. (a) How far must the lens be from the sensor array? (b) How tall is the image on the sensor array? Is it upright or inverted? Real or virtual? (c) An SLR digital camera often has pixels measuring \(8.0 \mu \mathrm{m} \times 8.0 \mu \mathrm{m}\). How many such pixels does the height of this image cover?

It's all done with mirrors. A photographer standing \(0.750 \mathrm{~m}\) in front of a plane mirror is taking a photograph of her image in the mirror, using a digital camera having a lens with a focal length of \(19.5 \mathrm{~mm}\). (a) How far is the lens from the light sensors of the camera? (b) If the camera is \(8.0 \mathrm{~cm}\) high, how high is its image on the sensors?

A \(135 \mathrm{~mm}\) telephoto lens for a \(35 \mathrm{~mm}\) camera has \(f\) -stops that range from \(f / 2.8\) to \(f / 22 .\) (a) What are the smallest and largest aperture diameters for this lens? What is the diameter at \(f / 11 ?\) (b) If a \(50 \mathrm{~mm}\) lens had the same \(f\) -stops as the telephoto lens, what would be the smallest and largest aperture diameters for that lens? (c) At a given shutter speed, what is the ratio of the greatest to the smallest light intensity of the film image? (d) If the shutter speed for correct exposure at \(f / 22\) is \(1 / 30 \mathrm{~s},\) what shutter speed is needed at \(f / 2.8 ?\) Calculate \(m_{1}\) and \(M_{2}\) for the two lenses and do not make the approximation that leads to Equation 25.4
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