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Chapter 34: Problem 28

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) Slide is 0.074 m from the lens. (b) Image is inverted. (c) Focal length is 0.071 m. (d) Lens is converging.

Step by step solution

01

Understand the Problem

We are given a slide, a lens, and a wall with a projected image. The image is 80 times larger than the slide. The distance from the slide to the wall is 6.00 m.
02

Use the Magnification Formula

Magnification is the ratio of the image height to the object height, and it is also given by \( \frac{d_i}{d_o} \). We know the magnification \( M = 80 \). Hence:\[ M = \frac{d_i}{d_o} = 80 \]where \( d_i \) is the image distance and \( d_o \) is the object distance (from the slide to the lens).
03

Relate Slide, Lens, and Wall Distances

The total distance from the slide to the wall is \( 6.00 \, \text{m} \). This distance is equal to the sum of the object distance \( d_o \) and the image distance \( d_i \). Thus:\[ d_o + d_i = 6.00 \, \text{m} \]
04

Solve for Object and Image Distances

We have two equations:1. \( \frac{d_i}{d_o} = 80 \)2. \( d_o + d_i = 6.00 \, \text{m} \)From the first equation, \( d_i = 80d_o \). Substitute in the second equation:\[ d_o + 80d_o = 6.00 \, \text{m} \]\[ 81d_o = 6.00 \, \text{m} \]\[ d_o = \frac{6.00}{81} \, \text{m} \approx 0.074 \, \text{m} \]
05

Calculate the Image Distance

Using \( d_i = 80d_o \) found in Step 4:\[ d_i = 80 \times 0.074 \, \text{m} = 5.92 \, \text{m} \]
06

Determine the Image Orientation

Since the magnification is positive and greater than 1, the image is enlarged and inverted with respect to the object. Thus, the image is inverted.
07

Apply Lens Formula to Find Focal Length

The lens formula is given by:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]Substitute the values of \( d_o \) and \( d_i \) to find \( f \):\[ \frac{1}{f} = \frac{1}{0.074} + \frac{1}{5.92} \]\[ f = \left( \frac{1}{0.074} + \frac{1}{5.92} \right)^{-1} \approx 0.071 \, \text{m} \]
08

Determine Lens Type

Since the focal length \( f \) is positive, the lens is a converging lens. Converging lenses can form real and inverted images, which is consistent with our findings.

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Key Concepts

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

Magnification Formula
In lens optics, magnification shows us how much larger or smaller the image is compared to the original object. It鈥檚 a simple ratio. This ratio compares the height of the image to the height of the object. Magnification can also tell us about distances. Use the formula:
  • \( M = \frac{d_i}{d_o} \)
Here, \( M \) is magnification, \( d_i \) is the image distance (distance from lens to image), and \( d_o \) is the object distance (distance from object to lens).
In our scenario, magnification is 80. This means the image is 80 times larger than the slide. Plugging this into the formula helps us understand the relationships between the various distances.
Lens Formula
The lens formula helps us find unknown distances or focal length. It鈥檚 an essential formula in optics and looks like this:
  • \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \)
Here, \( f \) is the focal length, \( d_o \) and \( d_i \) are the previously mentioned distances.
Using this relationship, if we know two of these values, we can find the third one. In our exercise, we needed to use known distances \( d_o \) and \( d_i \) to find the focal length \( f \). This concept is key in understanding how lenses focus light to form images.
Converging Lens
A converging lens, also known as a convex lens, bends light rays towards each other. Once the rays pass through, they converge or meet at a point.
These lenses are thicker in the center than at the edges. They are used for focusing light to a point or to produce a sharp image.
  • Converging lenses can form real images that are inverted when the object is outside the focal point.
  • The positive focal length of such a lens indicates its converging nature.
In our exercise, the positive focal length confirmed the lens used was indeed a converging lens.
Image Distance
Image distance \( d_i \) considers the space from the lens to where the image forms. This is crucial for locating where exactly the image will appear.
In optical setups, knowing the image distance helps us understand how the image will behave.
  • If an image appears far from the lens, it is usually larger compared to the object.
  • If close, the image may be smaller or the same size.
In our case, given the image size was 80 times larger and the total given distance (6 m), the image distance \( d_i \) was effectively calculated to be 5.92 m. This showed how far the image was projected onto the wall.

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Most popular questions from this chapter

A camera with a 90 -mm-focal-length lens is focused on an object 1.30 \(\mathrm{m}\) from the lens. To refocus on an object 6.50 \(\mathrm{m}\) from the lens, by how much must the distance between the lens and the film be changed? To refocus on the more distant object, is the lens moved toward or away from the film?

You hold a spherical salad bowl 90 \(\mathrm{cm}\) in front of your face with the bottom of the bowl facing you. The salad bowl is made of polished metal with a \(35-\mathrm{cm}\) radius of curvature. (a) Where is the image of your 2.0 -cm-tall nose located? (b) What are the image's size, orientation, and nature (real or virtual)?

A double-convex thin lens has surfaces with equal radii of curvature of magnitude \(2.50 \mathrm{cm} .\) Looking through this lens, you observe that it forms an image of a very distant tree at a distance of 1.87 \(\mathrm{cm}\) from the lens. What is the index of refraction of the lens?

The left end of a long glass rod 8.00 \(\mathrm{cm}\) in diameter, with an index of refraction 1.60 , is ground and polished to a convex hemispherical surface with a radius of 4.00 \(\mathrm{cm}\) . An object in the form of an arrow 1.50 \(\mathrm{mm}\) tall, at right angles to the axis of the rod, is located on the axis 24.0 \(\mathrm{cm}\) to the left of the vertex of the convex surface. Find the position and height of the image of the arrow formed by paraxial rays incident on the convex surface. Is the image erect or inverted?

You have a camera with a 35.0 -mm focal length lens and 36.0 -mm-wide film. You wish to take a picture of a 120 -m-long sailboat but find that the image of the boat fills only \(\frac{1}{4}\) of the width of the film. (a) How far are you from the boat? (b) How much closer must the boat be to you for its image to fill the width of the film?
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