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Chapter 34: Problem 30

A photographic slide is to the left of a lens. The lens projects an image of the slide onto a wall 6.00 \(\mathrm{m}\) to the right of the slide. The image is 80.0 times the size of the slide. (a) How far is the slide from the lens? (b) Is the image erect or inverted? (c) What is the focal length of the lens? (d) Is the lens converging or diverging?

Short Answer

Expert verified
(a) \(d_o \approx 0.0741 \text{ m}\), (b) inverted, (c) \(f \approx 0.0732 \text{ m}\), (d) converging lens.

Step by step solution

01

Understand the given information

We are given some distances and magnifications. Let's set them in symbols: Let the object distance be \(d_o\), the image distance be \(d_i = 6.00 \, \text{m} - d_o\) (since the image is 6.00 m away from the slide), and magnification \(m = -80.0\). The negative sign in magnification indicates the image is inverted.
02

Use the magnification formula

We use the magnification formula \( m = \frac{-d_i}{d_o} \). Substituting \(m = -80.0\), we have \(-80.0 = \frac{-(6.00 - d_o)}{d_o}\). Solve for \(d_o\) by cross-multiplying to get \(6.00 - d_o = 80.0 \times d_o\).
03

Solve for object distance \(d_o\)

Simplify the equation from Step 2: \(6.00 = 81d_o\). Divide both sides by 81 to find \(d_o\), giving \(d_o = \frac{6.00}{81} \approx 0.0741 \, \text{m}\).
04

Confirm the image is inverted

Since the magnification (\(m = -80.0\)) is negative, this means the image is inverted.
05

Use lens formula to find focal length

The lens formula is given by \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\). With \(d_i = 6.00 - d_o\) and \(d_o \approx 0.0741 \, \text{m}\), substitute these into the formula to find \(f\). Calculate \(\frac{1}{f} = \frac{1}{0.0741 \text{ m}} + \frac{1}{5.926 \text{ m}} \). Simplifying gives \(f \approx 0.0732 \text{ m}\).
06

Determine the type of lens

A diverging lens always forms images that are smaller, thus a magnification greater than 1 (80 in this case) implies it is a converging lens.

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Key Concepts

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

Image Magnification
Image magnification describes how much larger or smaller an image is compared to the actual object. In optics, magnification (\(m\)) is calculated using the formula \(m = \frac{-d_i}{d_o}\), where \(d_i\) is the image distance and \(d_o\) is the object distance. The sign of the magnification tells us the orientation of the image.
  • If \(m\) is positive, the image is upright.
  • If \(m\) is negative, the image is inverted.
In the given exercise, the magnification is \(-80.0\), meaning the image is 80 times larger than the object and inverted. Understanding magnification is vital to analyze whether you get upside-down images or images of different sizes.
Lens Formula
The lens formula relates the object distance \(d_o\), image distance \(d_i\), and the focal length \(f\) of a lens. It is expressed as \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\). This formula is crucial for solving questions involving lenses.

In the exercise, we used this formula to find the focal length after determining the object and image distances. Solving it requires the distances to be known, helping us easily calculate the characteristic, the focal length of the lens. The simplicity of the lens formula makes it a great tool for students, making it easier to solve complex optics problems by breaking down into these distances.
Focal Length
Focal length \(f\) is a measure of how strongly a lens converges or diverges light. It is the distance over which initially collimated rays are brought to a focus.
  • In converging lenses, the focal length is positive and is associated with converging light rays towards a focal point.
  • In diverging lenses, the focal length is negative and indicates that light rays are spreading outward.
In the context of the homework problem, we found the focal length using the lens equation, resulting in approximately \(0.0732 \text{m}\), signifying a converging lens.
Converging Lens
A converging lens, often called a convex lens, is thicker at the center than at the edges. It has the capability to bring parallel rays of light to a single focal point. These lenses are common in devices like cameras, glasses, and microscopes.

Its defining characteristic is having a positive focal length. Converging lenses can produce magnified images depending on the position of the object relative to the focal point. In the provided problem, the large positive magnification factor indicates the presence of a converging lens, emphasizing its utility in forming large, inverted images from relatively close objects.

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Most popular questions from this chapter

Dental Mirror. A dentist uses a curved mirror to view teeth on the upper side of the mouth. Suppose she wants an erect image with a magnification of 2.00 when the mirror is 1.25 \(\mathrm{cm}\) from a tooth. (Treat this problem as though the object and image lie along a straight line.) (a) What kind of mirror (concave or convex) is needed? Use a ray diagram to decide, without performing any calculations. (b) What must be the focal length and radius of curvature of this mirror? (c) Draw a principal-ray diagram to check your answer in part (b).

(a) Where is the near point of an eye for which a contact lens with a power of \(+2.75\) diopters is prescribed? (b) Where is the far point of an eye for which a contact lens with a power of \(-1.30\) diopters is prescribed for distant vision?

The radii of curvature of the surfaces of a thin converging meniscus lens are \(R_{1}=+12.0 \mathrm{cm}\) and \(R_{2}=+28.0 \mathrm{cm} .\) The index of refraction is 1.60 . (a) Compute the position and size of the image of an object in the form of an arrow 5.00 \(\mathrm{mm}\) tall, perpendicular to the lens axis, 45.0 \(\mathrm{cm}\) to the left of the lens. (b) A second converging lens with the same focal length is placed 3.15 \(\mathrm{m}\) to the right of the first. Find the position and size of the final image. Is the final image erect or inverted with respect to the original object? (c) Repeat part (b) except with the second lens 45.0 \(\mathrm{cm}\) to the right of the first.

A concave mirror has a radius of curvature of 34.0 \(\mathrm{cm} .\) (a) What is its focal length? (b) If the mirror is immersed in water (refractive index 1.33), what is its focal length?

cp calc You are in your car driving on a highway at 25 \(\mathrm{m} / \mathrm{s}\) when you glance in the passenger-side mirror (a convex mirror with radius of curvature 150 \(\mathrm{cm}\) ) and notice a truck approaching. If the image of the truck is approaching the vertex of the mirror at a speed of 1.9 \(\mathrm{m} / \mathrm{s}\) when the truck is 2.0 \(\mathrm{m}\) from the mirror, what is the speed of the truck relative to the highway?
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