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A converging lens has a focal length of \(88.00 \mathrm{cm}\). An object \(13.0 \mathrm{cm}\) tall is located \(155.0 \mathrm{cm}\) in front of this lens. (a) What is the image distance? (b) Is the image real or virtual? (c) What is the image height? Be sure to include the proper algebraic sign.

Short Answer

Expert verified
(a) Image distance is 203.66 cm. (b) The image is real. (c) Image height is -17.08 cm.

Step by step solution

01

Use Lens Formula

To find the image distance \(d_i\), we use the lens formula: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \] where \(f\) is the focal length, \(d_o\) is the object distance, and \(d_i\) is the image distance. The given values are \(f = 88.00\) cm and \(d_o = 155.0\) cm.
02

Rearrange Lens Formula

Rearrange the lens formula to solve for \(d_i\): \[ \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} \] Substitute the known values: \[ \frac{1}{d_i} = \frac{1}{88.00} - \frac{1}{155.0} \]
03

Calculate Image Distance

Calculate the right side of the equation: \[ \frac{1}{d_i} = 0.01136 - 0.00645 = 0.00491 \] Then calculate \(d_i\): \[ d_i = \frac{1}{0.00491} \approx 203.66 \text{ cm} \]
04

Determine Image Type

For a converging lens, if the image distance \(d_i\) is positive, the image is real. Since \(d_i = 203.66\) cm is positive, the image is real.
05

Use Magnification Formula for Image Height

The magnification formula is given by \(m = \frac{h_i}{h_o} = -\frac{d_i}{d_o}\). We need to find the image height \(h_i\), and we know \(h_o = 13.0\) cm. Substitute the known values into the formula: \[ m = -\frac{203.66}{155.0} \approx -1.314 \] \[ h_i = m \cdot h_o = -1.314 \cdot 13.0 \approx -17.08 \text{ cm} \]
06

Confirm Image Sign

The negative sign in the image height \(h_i\) suggests that the image is inverted, which is consistent with a real image formed by a converging lens.

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Key Concepts

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

Converging Lenses
A converging lens, also known as a convex lens, is thicker at the center than at the edges. This shape causes light rays passing through it to bend inward, or converge, toward a focal point. These lenses are commonly used in devices like cameras, glasses, and microscopes because they can focus light to form clear images.
In practice, when parallel light rays hit a converging lens, they are bent such that they meet at a specific point known as the focal point. This property makes converging lenses particularly useful for forming images. A real-world example where converging lenses play a critical role is in corrective eyeglasses, which help diverge or converge light rays to focus images on the retina.
Focal Length
The focal length of a lens is a crucial parameter that determines its converging power. In simple terms, it is the distance between the lens and its focal point where parallel light rays converge. For converging lenses, the focal length is positive because the real focus is on the opposite side of the light source.
Focal length not only dictates how strongly the lens converges light but also affects the size and position of the image formed. In mathematical terms, focal length is represented by the symbol \(f\). When using lenses in practical scenarios, knowing the focal length allows you to predict how the lens will behave in image formation.
Image Distance
Image distance, represented by \(d_i\), is the distance from the lens to the point where the image is formed. In lenses, this distance can provide insight into whether an image is real or virtual. If the image distance is positive, as calculated with a converging lens, the image formed is real and positioned on the opposite side to the object being observed.

Real images are typically inverted and can be projected onto a screen. In our exercise, we calculated an image distance of \(203.66\) cm, indicating a real and inverted image. Understanding how image distance works helps in designing systems where precise image placement is crucial, like in projectors or cameras.
Magnification Formula
The magnification formula provides a way to relate the size of an image to the size of the object. It is given by \(m = \frac{h_i}{h_o} = -\frac{d_i}{d_o}\), where:
  • \(m\) is the magnification.
  • \(h_i\) is the image height.
  • \(h_o\) is the object height.
  • \(d_i\) is the image distance.
  • \(d_o\) is the object distance.
Using this formula, you can determine how large or small an image will appear compared to the actual object. In the context of the exercise, the calculated magnification was \(-1.314\), indicating the image is inverted and slightly larger than the object. Negative magnification aligns with the idea that the image is inverted, a common characteristic of real images formed by converging lenses.

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Most popular questions from this chapter

Light is incident from air onto the surface of a liquid. The angle of incidence is \(53.0^{\circ},\) and the angle of refraction is \(34.0^{\circ} .\) At what angle of incidence would the reflected light be \(100 \%\) polarized?

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 object has an angular size of 0.0150 rad when placed at the near point \((21.0 \mathrm{cm})\) of an eye. When the eye views this object using a magnifying glass, the largest possible angular size of the image is 0.0380 rad. What is the focal length of the magnifying glass?

In a compound microscope, the focal length of the objective is \(3.50 \mathrm{cm}\) and that of the eyepiece is \(6.50 \mathrm{cm}\). The distance between the lenses is \(26.0 \mathrm{cm} .\) (a) What is the angular magnification of the microscope if the person using it has a near point of \(35.0 \mathrm{cm} ?\) (b) If, as usual, the first image lies just inside the focal point of the eyepiece (see Figure 26.32 ), how far is the object from the objective? (c) What is the magnification (not the angular magnification) of the objective?

The angular magnification of a telescope is 32800 times as large when you look through the correct end of the telescope as when you look through the wrong end. What is the angular magnification of the telescope?

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