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Chapter 26: Problem 62

An object is placed in front of a converging lens in such a position that the lens \((f=12.0 \mathrm{cm})\) creates a real image located \(21.0 \mathrm{cm}\) from the lens. Then, with the object remaining in place, the lens is replaced with another converging lens \((f=16.0 \mathrm{cm}) .\) A new, real image is formed. What is the image distance of this new image?

Short Answer

Expert verified
The new image distance is approximately 37.33 cm.

Step by step solution

01

Understand the Lens Formula

The lens formula relates the object distance \((d_o)\), the image distance \((d_i)\), and the focal length \((f)\) by the equation: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]. We have two cases: one with a focal length of 12 cm and another with a focal length of 16 cm.
02

Calculate Object Distance with First Lens

For the first lens with \(f = 12.0\, \text{cm}\) and \(d_i = 21.0\, \text{cm}\), use the lens formula to find \(d_o\):\[ \frac{1}{12} = \frac{1}{d_o} + \frac{1}{21} \]. Solve for \(\frac{1}{d_o}:\) \[ \frac{1}{d_o} = \frac{1}{12} - \frac{1}{21} = \frac{7 - 4}{84} = \frac{3}{84} = \frac{1}{28} \]. Hence, \(d_o = 28\, \text{cm}\).
03

Apply Object Distance to Second Lens

With the object distance \(d_o = 28\) cm from Step 2, use the second lens' focal length \(f = 16\) cm:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]. Plug in the values:\[ \frac{1}{16} = \frac{1}{28} + \frac{1}{d_i} \].
04

Solve for New Image Distance

Rearrange to find \(\frac{1}{d_i}\):\[ \frac{1}{d_i} = \frac{1}{16} - \frac{1}{28} \]. Convert to common denominators:\[ \frac{1}{d_i} = \frac{7 - 4}{112} = \frac{3}{112} \].Thus, \(d_i = \frac{112}{3} \approx 37.33\, \text{cm}\).

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Key Concepts

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

Converging Lens
A converging lens, also known as a convex lens, is a lens that brings parallel rays of light to a single point of focus. Its shape is thick in the middle and tapers at the edges. Converging lenses are used in various applications, such as cameras, glasses, and microscopes, to manipulate light and create real images.

When light passes through a converging lens, it bends inward. This bending occurs because the lens changes the direction of light rays, focusing them to a point. This focal point is where the image of an object is formed in front of or behind the lens.

A converging lens is often used in experiments and classrooms to demonstrate principles of optics. By understanding how these lenses work, one can predict how the image of an object will form based on its position relative to the lens's focal length.
Focal Length
The focal length of a lens is the distance from the lens's center to its focal point. It is a crucial parameter in determining how a lens forms an image. In the context of a converging lens, the focal length is the point where light rays converge.

Several factors affect the focal length:
  • Lens curvature: More curved lenses have shorter focal lengths.
  • Lens material: The refractive index of the lens material also impacts the focal length.
In our exercise, the lenses have focal lengths of 12 cm and 16 cm, which influence where the image forms. A smaller focal length results in a more powerful lens that converges light rays closer to the lens.

Using the lens formula, you can calculate the image or object distance if the other two values are known. This is important when designing optical systems to achieve desired image positions.
Image Distance
Image distance is the distance from the lens to the image formed by the lens. It is crucial in determining where the lamp or camera must be placed to capture the focused image clearly.

The lens formula connects image distance (\(d_i\)), object distance (\(d_o\)), and focal length (\(f\)) through:\(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\).
  • When the image distance is positive, a real image forms on the opposite side of the object.
  • If it's negative, an image forms on the same side as the object, signifying a virtual image.
In our exercise, the image distance changed when the lens was replaced, resulting from differing focal lengths. Calculating the image distance helps in adjusting the setup for precise focus. It is an essential part of configuring devices like cameras and projectors.

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Most popular questions from this chapter

A scuba diver, submerged under water, looks up and sees sunlight at an angle of \(28.0^{\circ}\) from the vertical. At what angle, measured from the vertical, does this sunlight strike the surface of the water?

A converging lens \(\left(f_{1}=24.0 \mathrm{cm}\right)\) is located \(56.0 \mathrm{cm}\) to the left of a diverging lens \(\left(f_{2}=-28.0 \mathrm{cm}\right)\). An object is placed to the left of the converging lens, and the final image produced by the two-lens combination lies \(20.7 \mathrm{cm}\) to the left of the diverging lens. How far is the object from the converging lens?

The near point of a naked eye is \(25 \mathrm{cm} .\) When placed at the near point and viewed by the naked eye, a tiny object would have an angular size of \(5.2 \times 10^{-5}\) rad. When viewed through a compound microscope, however, it has an angular size of \(-8.8 \times 10^{-3}\) rad. (The minus sign indicates that the image produced by the microscope is inverted.) The objective of the microscope has a focal length of \(2.6 \mathrm{cm},\) and the distance between the objective and the eyepiece is \(16 \mathrm{cm}\). Find the focal length of the eyepiece.

A ray of sunlight is passing from diamond into crown glass; the angle of incidence is \(35.00^{\circ} .\) The indices of refraction for the blue and red components of the ray are: blue \(\left(n_{\text {diamond }}=2.444, n_{\text {crown glass }}=1.531\right),\) and red \(\left(n_{\text {diamond }}=2.410, n_{\text {crown glass }}=1.520\right) .\) Determine the angle between the refracted blue and red rays in the crown glass.

An office copier uses a lens to place an image of a document onto a rotating drum. The copy is made from this image. (a) What kind of lens is used, converging or diverging? If the document and its copy are to have the same size, but are inverted with respect to one another, (b) how far from the document is the lens located and (c) how far from the lens is the image located? Express your answers in terms of the focal length \(f\) of the lens.
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