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An object with height \(4.00 \mathrm{~mm}\) is placed \(28.0 \mathrm{~cm}\) to the left of a converging lens that has focal length \(8.40 \mathrm{~cm} .\) A second lens is placed \(8.00 \mathrm{~cm}\) to the right of the converging lens. (a) What is the focal length of the second lens if the final image is inverted relative to the \(4.00-\mathrm{mm}\) -tall object and has height \(5.60 \mathrm{~mm} ?\) (b) What is the distance between the original object and the final image?

Short Answer

Expert verified
For part (a), the focal length of the second lens calculated in Step 4 is the answer. For part (b), the total distance calculated in Step 5 is the answer.

Step by step solution

01

Calculation for the first lens

First, we need to calculate the image distance from the first lens using the lens formula. We know that the object distance \(u = -28.0\) cm and the focal length \(f = 8.40\) cm. So, we rearrange the lens formula to solve for \(v\): \( 1/v = 1/f - 1/u\).
02

Calculate the magnification for the first lens

Next, we calculate the magnification of the image produced by the first lens using the formula \(m_1 = -v/u\), where \(v\) is the image distance calculated in step 1 and \(u = -28.0\) cm is the object distance.
03

Calculation for the second lens

Now we have found the object distance for the second lens. This is the image distance we calculated in Step 1 minus \(8.00\) cm (the distance to the second lens). The height of the object for the second lens will be the magnitude of the height produced by the first lens (\(h_o = |h_i| = |m_1| * 4.00 mm = |m_1| * 0.4\) cm). Using the given final magnification \(m_f = -5.60 mm / 4.00 mm = -1.4\), and knowing that the overall magnification is \(m_f = m_1 * m_2\), we can find the magnification produced by the second lens \(m_2 = m_f / m_1\).
04

Determine the focal length for the second lens

Once we have the magnification for the second lens and the object distance, we can calculate the image distance from the second lens using the magnification formula \(m_2 = -v/u\). Rearranging this gives \(v = -u * m_2\). We can then use the lens equation to solve for the focal length of the second lens \(f_2\).
05

Solve for the final image distance

The distance between the final image and the original object is the distance from the final image to the first lens (the image distance from the second lens) plus the original distance from the object to the first lens (28.0 cm)
06

Provide the calculated answers

The focal length of the second lens calculated in Step 4 is the answer for part (a). The total distance calculated in Step 5 is the answer for part (b).

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Key Concepts

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

Converging Lens
In geometric optics, a converging lens, also known as a convex lens, plays a pivotal role in focusing light. These lenses are thicker in the middle than at the edges, causing parallel light rays that pass through them to converge at a point known as the focal point. This focusing ability is what enables the lens to form images.

In practical applications, such as optical instruments, converging lenses can be used to magnify or focus images. They are essential in many devices, including cameras, eyeglasses, and microscopes. Understanding how these lenses manipulate light paths is fundamental in analyzing optical systems.

Some key properties of converging lenses include:
  • They have a positive focal length.
  • They can produce real images if the object is placed beyond the focal length.
  • They can produce virtual images if the object is placed within the focal length.
Knowing these properties assists in predicting image formation, which is crucial in solving optical problems.
Lens Formula
To understand how lenses focus light and form images, we use the lens formula. This formula provides the mathematical relationship between the object distance, the image distance, and the focal length of the lens.

The lens formula is given by:\[\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\]where:
  • \( f \) is the focal length of the lens,
  • \( v \) is the image distance (the distance from the lens to where the image forms),
  • \( u \) is the object distance (the distance from the object to the lens).
To use this formula, it's essential to note the sign convention: in optics, the object distance \( u \) is typically considered negative when the object is in front of the lens. This convention assists in accurately solving lens problems.

By rearranging the formula, we can solve for any of the unknowns if the others are known, facilitating the analysis of optical systems. This is a vital tool in understanding how lenses form images and is used extensively in exercises involving lenses.
Magnification
Another essential aspect of analyzing lenses is magnification. Magnification refers to how much larger or smaller an image is compared to the object itself. It is a vital concept in optics as it gives insight into the image's size and orientation relative to the object.

The formula for magnification \( m \) produced by a lens is:\[m = \frac{h_i}{h_o} = -\frac{v}{u}\]where:
  • \( h_i \) is the height of the image,
  • \( h_o \) is the height of the object,
  • \( m < 0 \) indicates an inverted image.
The negative sign in the magnification formula indicates that if the image is inverted relative to the object, the magnification will be negative. This inversion is crucial for understanding image characteristics in optical systems such as the one discussed in the problem.

Knowing how to calculate magnification helps determine the final image size and check the optical system's behavior. In practical scenarios, it lets users adjust devices to achieve the desired zoom or focus, making it a fundamental part of optical design and analysis.

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Most popular questions from this chapter

A concave mirror is to form an image of the filament of a headlight lamp on a screen \(8.00 \mathrm{~m}\) from the mirror. The filament is \(6.00 \mathrm{~mm}\) tall, and the image is to be \(24.0 \mathrm{~cm}\) tall. (a) How far in front of the vertex of the mirror should the filament be placed? (b) What should be the radius of curvature of the mirror?

When a camera is focused, the lens is moved away from or toward the digital image sensor. If you take a picture of your friend, who is standing \(3.90 \mathrm{~m}\) from the lens, using a camera with a lens with an \(85 \mathrm{~mm}\) focal length, how far from the sensor is the lens? Will the whole image of your friend, who is \(175 \mathrm{~cm}\) tall, fit on a sensor that is \(24 \mathrm{~mm} \times 36 \mathrm{~mm} ?\)

A converging lens with a focal length of \(12.0 \mathrm{~cm}\) forms a virtual image \(8.00 \mathrm{~mm}\) tall, \(17.0 \mathrm{~cm}\) to the right of the lens. Determine the position and size of the object. Is the image erect or inverted? Are the object and image on the same side or opposite sides of the lens? Draw a principal-ray diagram for this situation.

A spherical, concave shaving mirror has a radius of curvature of \(32.0 \mathrm{~cm}\). (a) What is the magnification of a person's face when it is \(12.0 \mathrm{~cm}\) to the left of the vertex of the mirror? (b) Where is the image? Is the image real or virtual? (c) Draw a principal-ray diagram showing the formation of the image.

Contact lenses are placed right on the eyeball, so the distance from the eye to an object (or image) is the same as the distance from the lens to that object (or image). A certain person can see distant objects well, but his near point is \(45.0 \mathrm{~cm}\) from his eyes instead of the usual \(25.0 \mathrm{~cm}\). (a) Is this person nearsighted or farsighted? (b) What type of lens (converging or diverging) is needed to correct his vision? (c) If the correcting lenses will be contact lenses, what focal length lens is needed and what is its power in diopters?

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