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A converging lens with a focal length of \(15.0 \mathrm{cm}\) and a diverging lens are placed \(25.0 \mathrm{cm}\) apart, with the converging lens on the left. A 2.00 -cm-high object is placed \(22.0 \mathrm{cm}\) to the left of the converging lens. The final image is \(34.0 \mathrm{cm}\) to the left of the converging lens. (a) What is the focal length of the diverging lens? (b) What is the height of the final image? (c) Is the final image upright or inverted?

Short Answer

Expert verified
(a) -35.47 cm, (b) -11.38 cm, (c) Inverted.

Step by step solution

01

Calculate the Image Distance of the First Lens

The object distance for the converging lens is \(d_o = 22.0 \ \mathrm{cm}\), and the focal length is \(f_1 = 15.0 \ \mathrm{cm}\). Using the lens formula, \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\), we solve for the image distance \(d_i\): \[ \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} = \frac{1}{15.0} - \frac{1}{22.0} \]\[ \frac{1}{d_i} = \frac{22.0 - 15.0}{15.0 \times 22.0} = \frac{7.0}{330.0} \] \[ d_i = \frac{330.0}{7.0} \approx 47.14 \ \mathrm{cm} \] Thus, the image formed by the first lens is \(47.14 \ \mathrm{cm}\) to the right of the converging lens.
02

Determine the Object Distance for the Second Lens

The diverging lens is placed \(25.0 \ \mathrm{cm}\) to the right of the converging lens. Therefore, the distance from the image formed by the first lens to the diverging lens is \(d = 47.14 \ \mathrm{cm} - 25.0 \ \mathrm{cm} = 22.14 \ \mathrm{cm}\). Since the image lies on the same side as the incoming light for the second lens, this distance becomes the object distance \(d_o\) for the diverging lens.
03

Apply the Lens Formula for the Diverging Lens

For the diverging lens, which eventually forms the final image at \(34.0 \ \mathrm{cm}\) to the left of the converging lens, its image distance relative to itself is \(d_i = - (34.0 \ \mathrm{cm} + 25.0 \ \mathrm{cm}) = -59.0 \ \mathrm{cm}\). Use the lens formula again to find the focal length \(f_2\): \[ \frac{1}{f_2} = \frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{22.14} + \frac{1}{-59.0} \] Solve to find \[ \frac{1}{f_2} = \frac{59.0 - 22.14}{22.14 \times -59.0} = \frac{36.86}{-1307.26} \] \[ f_2 = \frac{-1307.26}{36.86} \approx -35.47 \ \mathrm{cm}\] The focal length of the diverging lens is \(-35.47 \ \mathrm{cm}\).
04

Calculate the Magnification and Height of the Final Image

The magnification \(m_1\) of the first lens is \(m_1 = -\frac{d_i}{d_o} = -\frac{47.14}{22.0} \approx -2.14\). The magnification \(m_2\) of the second lens is \(m_2 = -\frac{d_i}{d_o} = -\frac{-59.0}{22.14} \approx 2.66\). The total magnification \(m_{total}\) is the product of both: \(m_{total} = m_1 \times m_2 = -2.14 \times 2.66 = -5.69\). The height of the final image \(h_i\) is \(h_i = m_{total} \times h_o = -5.69 \times 2.0 \approx -11.38 \ \mathrm{cm}\).
05

Determine the Orientation of the Final Image

Since the total magnification \(m_{total}\) is negative, the image is inverted. Thus, the final image is inverted.

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Key Concepts

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

Lens formula
Understanding the lens formula is crucial for analyzing how lenses create images. The lens formula is defined as \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\), where \(f\) is the focal length, \(d_o\) is the object distance, and \(d_i\) is the image distance. This formula helps us relate these distances and determine the nature of the image formed by a lens. Whether dealing with a converging or diverging lens, the lens formula will guide you through calculating where the image will form.
For converging lenses, which are thicker at the center than at the edges, they have positive focal lengths, while diverging lenses, which are thinner at the center, have negative focal lengths. This aspect is crucial when applying the lens formula to solve problems around image formation and lens properties.
Image formation
Image formation by lenses depends on the properties of the lenses used and the placement of the object relative to them. When light passes through a lens, it can converge or diverge to form images at specific distances.
Lenses create images through refraction. Refracted light rays converge or diverge at points which are either real or virtual images. Real images are formed when light rays actually meet, whereas virtual images appear to meet when extended backwards. The type of image depends on the type of lens and its position relative to the object.
  • If the image distance \(d_i\) is positive, the image is real and formed on the opposite side of the lens from the object.
  • If \(d_i\) is negative, it is a virtual image, appearing on the same side of the lens as the object.
Using the lens formula, you can determine the image distance and identify whether the image is real or virtual.
Converging lens
A converging lens, usually convex, bends light rays towards a focal point. It is thicker in the middle and has a positive focal length. These lenses are commonly used in applications where image magnification is needed, like cameras or glasses for hyperopia (farsightedness).
The process of forming images with a converging lens often involves determining where the light rays will meet to form an image. This requires using the lens formula to find the image distance (\(d_i\)) given the object distance (\(d_o\)) and focal length (\(f\)).
  • Converging lenses create real images if the object is placed outside the focal point.
  • If the object is inside the focal point, the lens produces a virtual image.
Understanding how light bends with converging lenses lets us predict image properties like size and orientation.
Diverging lens
Diverging lenses, often concave, spread out light rays, making them appear to diverge from a common point. These lenses are thinner in the middle and have a negative focal length. They are used in glasses for myopia (nearsightedness) as they help to spread out light before it reaches the eye.
Image formation with a diverging lens is different from a converging lens. Light diverges after passing through the lens, forming a virtual image on the same side as the object. The image is usually smaller and upright compared to the object.
  • In calculations, the focal length (\(f\)) is considered negative, reflecting the lens's diverging property.
  • Diverging lenses always produce virtual images that are erect relative to the object.
Using the lens formula helps determine the precise position and size of the virtual image formed by a diverging lens.

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Most popular questions from this chapter

A convex lens of power +12 D is used as a magnifier to examine a wildflower. What is the angular magnification if the final image is at (a) infinity or (b) the near point of \(25 \mathrm{cm} ?\)
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