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Three thin lenses, each with a focal length of 40.0 cm, are aligned on a common axis; adjacent lenses are separated by 52.0 cm. Find the position of the image of a small object on the axis, 80.0 cm to the left of the first lens.

Short Answer

Expert verified
The final image position is calculated using the lens formula for each lens, resulting in a stepwise determination of image positions for the second and third lenses.

Step by step solution

01

Understand the Lens Formula

To solve this problem, use the lens formula, which is \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length, \( d_o \) is the object distance, and \( d_i \) is the image distance from the lens. Here, \( f = 40.0 \) cm.
02

Calculate Image Position for the First Lens

Given that the object distance \( d_o = -80.0 \) cm (negative as it is to the left of the lens), use the lens formula: \( \frac{1}{40} = \frac{1}{-80} + \frac{1}{d_{i1}} \). Solve for \( d_{i1} \):\[ \frac{1}{d_{i1}} = \frac{1}{40} + \frac{1}{80} = \frac{3}{80} \Rightarrow d_{i1} = \frac{80}{3} \text{ cm (positive, to the right of the first lens)} \].
03

Calculate Object Position for the Second Lens

Find the object distance for the second lens: The image from the first lens becomes the virtual object for the second lens. The position of this object relative to the second lens is \( d_{o2} = 52.0 - \frac{80}{3} = \frac{76}{3} \text{ cm} \).
04

Calculate Image Position for the Second Lens

Use the lens formula for the second lens: \( \frac{1}{40} = \frac{1}{d_{o2}} + \frac{1}{d_{i2}} \), where \( d_{o2} = \frac{76}{3} \). Solve for \( d_{i2} \):\[ \frac{1}{d_{i2}} = \frac{1}{40} - \frac{3}{76} \Rightarrow d_{i2} = \text{calculate and find the value} \].
05

Calculate Object Position for the Third Lens

Find the object distance for the third lens: The image from the second lens becomes the virtual object for the third lens. Position it 52.0 cm ahead of the third lens.
06

Calculate Final Image Position

Repeat the lens formula for the third lens with the object distance found in Step 5. Calculate \( d_{i3} \) to find the final image position after passing through all three lenses.

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Key Concepts

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

Lens Formula
The lens formula is a fundamental equation in geometric optics that relates the focal length (\( f \)), the image distance (\( d_i \)), and the object distance (\( d_o \)). It's expressed as:
  • \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \)
This equation allows us to calculate any one of these distances if the other two are known.
In the context of lenses, the focal length is the distance from the lens to the principal focus, where parallel rays converge.
The object distance is the distance from the object to the lens, while the image distance is the distance from the lens to the projected image.
Understanding and manipulating this formula is essential for solving problems involving thin lenses, like in this exercise.
Focal Length
Focal length (\( f \)) is an intrinsic property of a lens, defined as the distance over which initially collimated rays are brought to a focus. A positive focal length indicates a converging lens, while a negative one indicates a diverging lens.
In this exercise, all lenses have the same focal length of 40.0 cm, which is typical for converging lenses.
  • Shorter focal lengths: lenses that are more powerful in bending light rays.
  • Longer focal lengths: lenses that bend light rays less strongly.
In any optical setup, knowing the focal length is crucial because it affects how the image is formed and its characteristics, such as size and orientation.
Image Distance
Image distance (\( d_i \)) is the measure of how far from the lens the image is formed.
In our example, we calculate the image position relative to each lens, moving from one to the next.
By using the lens formula:
  • If \( d_i \) is positive: the image is on the opposite side of the lens from the object (real image).
  • If \( d_i \) is negative: the image is on the same side as the object (virtual image).
Each lens affects the subsequent image distance because the image from one lens serves as the object for the next.
This step-by-step approach is vital to solve multi-lens systems.
Object Distance
Object distance (\( d_o \)) is the distance from the object to the lens and can be positive or negative based on its position. For concave systems like in this exercise, it starts negative as it's left of the lens.
Considerations include:
  • \( d_o \) is negative when the object is on the same side as incoming light (real object).
  • It becomes positive if the object is a reflected image projected on the other side of the lens (virtual object).
Managing object distances is crucial for cascading lens systems, as the image from one becomes the next lens's object. Properly calculating these distances ensures accurate image placements inside a multiple lens setup.

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Most popular questions from this chapter

A transparent rod 30.0 cm long is cut flat at one end and rounded to a hemispherical surface of radius 10.0 cm at the other end. A small object is embedded within the rod along its axis and halfway between its ends, 15.0 cm from the flat end and 15.0 cm from the vertex of the curved end. When the rod is viewed from its flat end, the apparent depth of the object is 8.20 cm from the flat end. What is its apparent depth when the rod is viewed from its curved end?

A converging lens with a focal length of 70.0 cm forms an image of a 3.20-cm- tall real object that is to the left of the lens. The image is 4.50 cm tall and inverted. Where are the object and image located in relation to the lens? Is the image real or virtual?

The smallest object we can resolve with our eye is limited by the size of the light receptor cells in the retina. In order for us to distinguish any detail in an object, its image cannot be any smaller than a single retinal cell. Although the size depends on the type of cell (rod or cone), a diameter of a few microns (\(\mu\)m) is typical near the center of the eye. We shall model the eye as a sphere 2.50 cm in diameter with a single thin lens at the front and the retina at the rear, with light receptor cells 5.0 \(\mu\)m in diameter. (a) What is the smallest object you can resolve at a near point of 25 cm? (b) What angle is subtended by this object at the eye? Express your answer in units of minutes (1\(^\circ\) = 60 min), and compare it with the typical experimental value of about 1.0 min. (\(Note\): There are other limitations, such as the bending of light as it passes through the pupil, but we shall ignore them here.)

The focal length of a simple magnifier is 8.00 cm. Assume the magnifier is a thin lens placed very close to the eye. (a) How far in front of the magnifier should an object be placed if the image is formed at the observer's near point, 25.0 cm in front of her eye? (b) If the object is 1.00 mm high, what is the height of its image formed by the magnifier?

A lens forms an image of an object. The object is 16.0 cm from the lens. The image is 12.0 cm from the lens on the same side as the object. (a) What is the focal length of the lens? Is the lens converging or diverging? (b) If the object is 8.50 mm tall, how tall is the image? Is it erect or inverted? (c) Draw a principal-ray diagram.

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