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\(26.3\) at reviews the concepts that play a role in this problem. 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) The image distance is approximately 175.14 cm. (b) The image is real. (c) The image height is approximately -14.68 cm.

Step by step solution

01

Understanding the Lens Formula

The lens formula relates the object distance \(d_o\), the image distance \(d_i\), and the focal length \(f\) of a lens. It is given by: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]
02

Substituting Known Values to Find Image Distance

We know that \(f = 88.00 \text{ cm}\) and the object distance \(d_o = 155.0 \text{ cm}\). We need to find \(d_i\). Substitute these values into the lens formula: \[ \frac{1}{88.00} = \frac{1}{155.0} + \frac{1}{d_i} \] After rearranging the formula, solve for \(\frac{1}{d_i}\): \[ \frac{1}{d_i} = \frac{1}{88.00} - \frac{1}{155.0} = \frac{0.01136}{\text{approximate value}} \] This results in \(d_i \approx 175.14 \text{ cm}\).
03

Determining the Nature of the Image

An image is real if the image distance \(d_i\) is positive, and virtual if \(d_i\) is negative. Since \(d_i = 175.14 \text{ cm}\) is positive, the image is real.
04

Finding the Image Height

Use the magnification formula, which relates the object height \(h_o\), the image height \(h_i\), the object distance \(d_o\), and the image distance \(d_i\). The formula is given by: \[ m = \frac{h_i}{h_o} = -\frac{d_i}{d_o} \] Given \(h_o = 13.0 \text{ cm}\), \(d_i = 175.14 \text{ cm}\), and \(d_o = 155.0 \text{ cm}\), substitute these into the magnification formula to solve for \(h_i\): \[ \frac{h_i}{13.0} = -\frac{175.14}{155.0} \] Hence, \(h_i = -\frac{175.14}{155.0} \cdot 13.0 \approx -14.68 \text{ cm}\). The negative sign indicates that the 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
The lens formula is crucial for understanding how images are formed by lenses. It is a mathematical equation that relates the focal length of the lens, the distance from the object to the lens, and the distance from the image to the lens. The formula is written as: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]Where:
  • \( f \) is the focal length of the lens.
  • \( d_o \) is the distance from the object to the lens.
  • \( d_i \) is the distance from the image to the lens.

This formula is derived from the principle of optical lenses, where light converges or diverges to form images. It is essential to identify the focal length correctly, which is positive for converging lenses and negative for diverging lenses.
Understanding this formula helps in calculating other significant properties of images, such as their size and orientation.
Image Distance
Image distance is an important concept in understanding how lenses work. It refers to how far the formed image is from the lens. In the context of the lens formula, the image distance \( d_i \) can be calculated if the focal length \( f \) and object distance \( d_o \) are known.
In solving for \( d_i \), we frequently rearrange the lens formula:\[ \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} \]By solving this equation, the value of \( d_i \) can be found. In our example, with a focal length of \( 88.00 \) cm and an object distance of \( 155.0 \) cm, \( d_i \) came out to approximately \( 175.14 \) cm.
Importantly, the sign of \( d_i \) reveals more than just where the image is located. A positive \( d_i \) indicates a real image, which means it forms on the opposite side of the lens compared to where the object is placed. Conversely, when \( d_i \) is negative, it represents a virtual image.
Magnification
Magnification describes how much larger or smaller the image is compared to the actual object. This is an essential factor when analyzing optical lenses because it allows us to predict the size and orientation of the image. The magnification \( m \) is given by:\[ m = \frac{h_i}{h_o} = -\frac{d_i}{d_o} \]Where:
  • \( h_i \) is the height of the image.
  • \( h_o \) is the height of the object.
  • \( d_i \) and \( d_o \) are image and object distances, respectively.

For our lens example:- The object was \( 13.0 \) cm tall, creating an image with a height calculated to be approximately \( -14.68 \) cm using the formula. - The negative magnification value indicates that the image is inverted.
Magnification can significantly affect what we perceive through optical lenses, whether in microscopes, glasses, or cameras. Understanding this allows us to adjust and modify lens systems for desired outcomes.

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Most popular questions from this chapter

Violet light and red light travel through air and strike a block of plastic at the same angle of incidence. The angle of refraction is \(30.400^{\circ}\) for the violet light and \(31.200^{\circ}\) for the red light. The index of refraction for violet light in plastic is greater than that for red light by \(0.0400 .\) Delaying any rounding off of calculations until the very end, find the index of refraction for violet light in plastic.

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?

An object is located \(30.0 \mathrm{~cm}\) to the left of a converging lens whose focal length is \(50.0 \mathrm{~cm} .\) (a) Draw a ray diagram to scale and from it determine the image distance and the magnification, (b) Use the thin-lens and magnification equations to verify your answers to part (a).

The contacts wom by a farsighted person allow her to see objects clearly that are as close as \(25.0 \mathrm{~cm}\), even though her uncorrected near point is \(79.0 \mathrm{~cm}\) from her eyes. When she is looking at a poster, the contacts form an image of the poster at a distance of \(217 \mathrm{~cm}\) from her eyes. (a) How far away is the poster actually located? (b) If the poster is \(0.350 \mathrm{~m}\) tall, how tall is the image formed by the contacts?

An astronomical telescope has an angular magnification of -184 and uses an objective with a focal length of \(48.0 \mathrm{~cm}\). What is the focal length of the eyepiece?

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