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An object \(2.00 \mathrm{cm}\) high is placed \(40.0 \mathrm{cm}\) to the left of a converging lens having a focal length of \(30.0 \mathrm{cm} .\) A diverging lens with a focal length of \(-20.0 \mathrm{cm}\) is placed \(110 \mathrm{cm}\) to the right of the converging lens. (a) Determine the position and magnification of the final image. (b) Is the image upright or inverted? (c) What If? Repeat parts (a) and (b) for the case where the second lens is a converging lens having a focal length of \(+20.0 \mathrm{cm}\).

Short Answer

Expert verified
For (a) and (b), the position of the final image and the magnification will be calculated using lens formula and multiplying the magnifications derived from each lens step. Orientation of the image will be upright if final magnification is positive and inverted if it is negative. For (c), the entire process will be repeated but the focal length of second lens would be considered as converging.

Step by step solution

01

- Find first intermediate image

Use the lens formula, \(1/f=1/v-1/u\), where 'f' is the focal length, 'v' is the image distance and 'u' is the object distance. Substitute 'f' with \(30.0 \mathrm{cm}\) (focal length of the converger), 'u' with \(-40.0 \mathrm{cm}\) (distance of the object from lens – it is negative as per Cartesian sign convention). Solve this equation to find 'v', which will give the location of the image formed by first lens relative to it. Also find magnification for this step, \(m1 = -v/u\)
02

- Find final image using the second lens

The image from step 1 becomes object for the next lens with a new object distance being the distance between the two lenses minus the absolute of first image distance. Now apply the lens formula again for the diverging lens having a focal length of \(-20.0 \mathrm{cm}\). Once again find the new image distance and magnification \(m2 = -v/u\)
03

- Determine the final magnification and position of final image

Multiply the magnifications from step 1 and step 2 to get the final overall magnification, \(m = m1 * m2\). The overall position of final image would be the distance between lenses and image distance from step 2.
04

- Determine the orientation of the image

If final magnification is positive, the image would be upright else it would be inverted.
05

- What If scenario

Repeat steps 1 through 4 but change the focal length of the second lens to \(20.0 \mathrm{cm}\)

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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 designed to bring light rays to a point. This is achieved through its curved surface. The main feature is that it can focus light, making objects appear larger and sharp.
Converging lenses are defined by their focal length, which is the distance from the lens's center to the focal point where light rays meet. This focal length is positive due to the nature of the lens
  • Positive focal length
  • Focal point brings light together
  • Used in a variety of optical devices like cameras and glasses

In the problem, the converging lens has a focal length of \(30.0 \; \mathrm{cm}\). Understanding how this lens works is crucial for determining how an image is formed. The object placed at \(40.0 \; \mathrm{cm}\) to the left of the lens indicates the distance from where light begins its course through the lens.
Diverging Lens
Unlike converging lenses, diverging lenses (or concave lenses) spread light rays apart. This means they cause parallel light to diverge as if it originated from a focal point behind the lens.
Diverging lenses have a negative focal length, representing this outward spread of light.
  • Negative focal length
  • Light is spread out
  • Used in applications like correcting short-sightedness

In the textbook problem, the diverging lens has a focal length of \(-20.0 \; \mathrm{cm}\), situated \(110 \; \mathrm{cm}\) to the right of the converging lens. This divergence will affect the journey of the light rays coming from the image formed by the first lens, changing its position and size.
Image Magnification
The concept of image magnification is central to understanding how lenses work. It refers to the factor by which an image is enlarged compared to the object. This ratio is a result of the distances and focal lengths involved.
To calculate magnification, the formula \(m = -\frac{v}{u}\) is used, where \(v\) is the image distance from the lens and \(u\) is the object distance from the lens.
  • Magnification less than 1 means the image is smaller than the object
  • Greater than 1 means the image is larger
  • Positive magnification indicates an upright image
  • Negative magnification indicates an inverted image

In the problem's solution steps, initial magnification is calculated after the converging lens. Then, in combination with the diverging lens, the final magnification is determined, and the orientation can be inferred through its sign.
Lens Formula
The lens formula is a mathematical representation that relates the object distance, image distance, and the focal length of a lens. It is crucial for finding the precise location of an image formed by the lens.
The formula is written as: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]Where:
  • \(f\) is the focal length of the lens (positive for converging, negative for diverging)
  • \(v\) is the image distance
  • \(u\) is the object distance
This formula allows you to solve for one variable if the others are known. Applying the lens formula in step-by-step solutions enables finding the positions of intermediate and final images in optics problems.

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Most popular questions from this chapter

A thin lens has a focal length of \(25.0 \mathrm{cm} .\) Locate and describe the image when the object is placed (a) \(26.0 \mathrm{cm}\) and (b) \(24.0 \mathrm{cm}\) in front of the lens.

The accommodation limits for Nearsighted Nick's eyes are \(18.0 \mathrm{cm}\) and \(80.0 \mathrm{cm} .\) When he wears his glasses, he can see faraway objects clearly. At what minimum distance is he able to see objects clearly?

An object is located \(20.0 \mathrm{cm}\) to the left of a diverging lens having a focal length \(f=-32.0 \mathrm{cm} .\) Determine (a) the location and (b) the magnification of the image. (c) Construct a ray diagram for this arrangement.

In some types of optical spectroscopy, such as photoluminescence and Raman spectroscopy, a laser beam exits from a pupil and is focused on a sample to stimulate electromagnetic radiation from the sample. The focusing lens usually has an antireflective coating preventing any light loss. Assume a \(100-\mathrm{mW}\) laser is located \(4.80 \mathrm{m}\) from the lens, which has a focal length of \(7.00 \mathrm{cm} .\) (a) How far from the lens should the sample be located so that an image of the laser exit pupil is formed on the surface of the sample? (b) If the diameter of the laser exit pupil is \(5.00 \mathrm{mm},\) what is the diameter of the light spot on the sample? (c) What is the light intensity at the spot?

A dedicated sports car enthusiast polishes the inside and outside surfaces of a hubcap that is a section of a sphere. When she looks into one side of the hubcap, she sees an image of her face \(30.0 \mathrm{cm}\) in back of the hubcap. She then flips the hubcap over and sees another image of her face \(10.0 \mathrm{cm}\) in back of the hubcap. (a) How far is her face from the hubcap? (b) What is the radius of curvature of the hubcap?

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