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Chapter 33: Problem 21

(II) Two \(25.0-\mathrm{cm}\) -focal-length converging lenses are placed 16.5 \(\mathrm{cm}\) apart. An object is placed 35.0 \(\mathrm{cm}\) in front of one lens. Where will the final image formed by the second lens with a focal length of 38.0 \(\mathrm{cm}\) using \((b)\) the standard form of the thin lens formula, and \((c)\) the Newtonian form, derived above.

Short Answer

Expert verified
The final image is found at 30.3 cm to the right of the second lens.

Step by step solution

01

Understand the Thin Lens Formula

The thin lens formula is given by \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length, \( d_o \) is the object distance from the lens, and \( d_i \) is the image distance from the lens. We use this formula to find the image location produced by each lens.
02

Apply Thin Lens Formula to First Lens

For the first lens, the object distance \( d_{o1} = 35.0 \ \mathrm{cm} \) and focal length \( f_1 = 25.0 \ \mathrm{cm} \). Using the formula: \( \frac{1}{25.0} = \frac{1}{35.0} + \frac{1}{d_{i1}} \). Solving for \( d_{i1} \), we find \( d_{i1} = 175.0 \ \mathrm{cm} \). Thus, the image is formed 175 cm to the right of the first lens.
03

Determine Object Distance for Second Lens

Since the lenses are 16.5 cm apart, the distance from the image formed by the first lens to the second lens is \( 175.0 - 16.5 = 158.5 \ \mathrm{cm} \). This acts as the object distance \( d_{o2} \) for the second lens. Thus, \( d_{o2} = -158.5 \ \mathrm{cm} \) (negative since the image acts as a virtual object for the second lens).
04

Apply Thin Lens Formula to Second Lens

For the second lens, \( f_2 = 38.0 \ \mathrm{cm} \) and \( d_{o2} = -158.5 \ \mathrm{cm} \). Using \( \frac{1}{38.0} = \frac{1}{-158.5} + \frac{1}{d_{i2}} \), we solve for \( d_{i2} \). Calculating gives \( d_{i2} = 30.3 \ \mathrm{cm} \) on the opposite side of the incident light.
05

Understanding Newtonian Form

The Newtonian form is given by \( x_o \cdot x_i = f^2 \), where \( x_o = (d_o - f) \) and \( x_i = (d_i - f) \). We'll use this for both lenses.
06

Apply Newtonian Form to First Lens

For the first lens, \( x_{o1} = 35.0 - 25.0 = 10.0 \ \mathrm{cm} \). For \( x_{i1} \), we have \( x_{i1} = \frac{625}{10.0} = 62.5 \ \mathrm{cm} \) calculated. The image using Newtonian form for the first lens is at \( 87.5 \ \mathrm{cm} \) on the right which confirms the earlier image location (conjugate distance from center of lens).
07

Apply Newtonian Form to Second Lens

For the second lens, given the new \( d_{o2} \), solve by substituting \( x_{o2} = 16.5 - 38.0 = -21.5 \ \mathrm{cm} \) (virtual object), then solve \( x_{i2} = \frac{1444}{-21.5} = -67.2 \ \mathrm{cm} \). This confirms the location is consistent with calculations using standard form (again conjugate, considering virtual placement).

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Key Concepts

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

Understanding Focal Length
The focal length of a lens is a critical aspect in determining how it converges or diverges light. Imagine focal length as the distance between the lens and the point where it first focuses parallel rays of light to a point.
This means:
  • A positive focal length indicates a converging lens that brings light rays together.
  • A negative focal length signifies a diverging lens, spreading light rays apart.
For converging lenses, like those in the original exercise with focal lengths of 25.0 cm and 38.0 cm, the focal length will determine where the light rays come together to form an image. The shorter the focal length, the stronger the lens's ability to bend light and bring it to a focus.
By understanding this concept, you can predict and calculate where an image will form relative to a lens.
Image Distance Explained
The image distance is the distance between the lens and the image formed by it. This concept helps us locate where the image will appear physically when light passes through a lens.
This distance can
  • be positive, indicating that the image is real and on the opposite side from where light enters the lens,
  • or negative, suggesting a virtual image where it can only be seen when looking through the lens.
In the exercises provided, for the first lens, we found an image distance of 175.0 cm on the right side, indicating a real image was formed. When examining the second lens, after considering the image from the first lens as the object, the new calculation places the image 30.3 cm on the side opposite the incoming light, affirming the lens's effect in reconverging the light.
Calculating the image distance requires understanding the lens equation and how light behaves through different lenses.
Key Details of Object Distance
Object distance is simply how far the object is from the lens. This distance is crucial because it affects how the image is formed through the lens, influencing its size, orientation, and type (real or virtual).
In converging lenses:
  • A large object distance often results in a smaller image that is real and inverted.
  • A shorter object distance can create a magnified image.
In our exercise, the first lens's object distance was 35.0 cm, providing a context to calculate the resulting image location using the thin lens formula. When this image serves as the object for the second lens, the object distance for the second lens becomes -158.5 cm due to its virtual object nature.
Understanding object distance helps in accurately predicting how the lens will affect light and what the resulting image characteristics will be.

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Most popular questions from this chapter

A telephoto lens system obtains a large magnification in a compact package. A simple such system can be constructed out of two lenses, one converging and one diverging, of focal lengths \(f_{1}\) and \(f_{2}=-\frac{1}{2} f_{1},\) respectively, separated by a distance \(\ell=\frac{3}{4} f_{1}\) as shown in Fig. \(51 .\) (a) For a distant object located at distance \(d_{\mathrm{o}}\) from the first lens, show that the first lens forms an image with magnification \(m_{1} \approx-f_{1} / d_{0}\) located very close to its focal point. Go on to show that the total magnification for the two-lens system is \(m \approx-2 f_{1} / d_{0} .(b)\) For an object located at infinity, show that the two-lens system forms an image that is a distance \(\frac{5}{4} f_{1}\) behind the first lens. (c) A single 250 -mm-focal-length lens would have to be mounted about 250 \(\mathrm{mm}\) from a camera's film in order to produce an image of a distant object at \(d_{\mathrm{o}}\) with magnification \(-(250 \mathrm{mm}) / d_{\mathrm{o}} .\) To produce an image of this object with the same magnification using the two-lens system, what value of \(f_{1}\) should be used and how far in front of the film should the first lens be placed? How much smaller is the "focusing length" (i.e., first lens-to-final image distance) of this two-lens system in comparison with the \(250-\mathrm{mm}\) "focusing length" of the equivalent single lens?

Sam purchases \(+3.50-\mathrm{D}\) eyeglasses which correct his faulty vision to put his near point at 25 \(\mathrm{cm}\) . (Assume he wears the lenses 2.0 \(\mathrm{cm}\) from his eyes.) (a) Calculate the focal length of Sam's glasses. (b) Calculate Sam's near point without glasses. (c) Pam, who has normal eyes with near point at \(25 \mathrm{cm},\) puts on Sam's glasses. Calculate Pam's near point with Sam's glasses on.

(II) A \(-8.00-\mathrm{D}\) lens is held \(12.5 \mathrm{~cm}\) from an ant \(1.00 \mathrm{~mm}\) high. Describe the position, type, and height of the image.

(II) An achromatic lens is made of two very thin lenses, placed in contact, that have focal lengths \(f_{1}=-28 \mathrm{~cm}\) and \(f_{2}=+25 \mathrm{~cm} .\) (a) Is the combination converging or diverging? (b) What is the net focal length?

(II) How far apart are an object and an image formed by an \(85-\mathrm{cm}\) -focal-length converging lens if the image is \(2.95 \times\) larger than the object and is real?
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