Learning Goal: To learn the quantitative use of the lens equation, as well as how to determine qualitative properties of solutions. In working with lenses, there are three important quantities to consider. The object distance is the distance along the axis of the lens to the object. The image distance s' is the distance along the axis of the lens to the image. The focal length f is an intrinsic property of the lens. These three quantities are related through the equation 1 + 1 = 1 Note that this equation is valid only for thin, spherical lenses. Unless otherwise specified, a lens problem always assumes that you are using thin, spherical lenses. The equation above allows you to calculate the locations of images and objects. Frequently. you will also be interested in the size of the image or object, particularly if you are considering a magnifying glass or microscope. The ratio of the size of an image to the size of the object is called the magnification. It is given by m = 4 where y' is the height of the image and y is the height of the object. The second equality allows you to find the size of the image (or object) with the information provided by the thin lens equation All of the quantities in the above equations can take both positive and negative values. Positive distances correspond to real images or objects, while negative distances correspond to virtual images or objects. Positive heights correspond to upright images or objects, while negative heights correspond to inverted images or objects. The following table summarizes these properties: positive negative virtual virtual 8 real real y upright inverted upright invented The focal length f can also be positive or negative. A positive focal length corresponds to a converging lens, while a negative focal length corresponds to a diverging lens. Consider an object with a 12 cm that produces an image with 15 cm. Note that whenever you are working with a physical object, the object distance will be positive (in multiple optics setups, you will encount "objects that are actually images, but that is not a possibility in this problem). A positive image distance means that the image is formed on the side of the lens from which the light emerges ▾ Part A ▾ Find the focal length of the lens that produces the image described in the problem introduction using the thin lens equation Express your answer in centimeters, as a fraction or to three significant figures. f = 8.67 cm ✓ Correct Part B Previous Answers Considering the sign of f. is the lens converging or diverging? converging Ⓒdiverging Part C Previous Anewere ✓ Correct

College Physics
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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i need help with part I

 

Learning Goal:
To learn the quantitative use of the lens equation, as well as how to determine qualitative
properties of solutions.
In working with lenses, there are three important quantities to consider. The object distance
8 is the distance along the axis of the lens to the object. The image distance s' is the
distance along the axis of the lens to the image. The focal length f is an intrinsic property
of the lens. These three quantities are related through the equation
= + = = }
Note that this equation is valid only for thin, spherical lenses. Unless otherwise specified, a
lens problem always assumes that you are using thin, spherical lenses.
The equation above allows you to calculate the locations of images and objects. Frequently,
you will also be interested in the size of the image or object, particularly if you are
considering a magnifying glass or microscope. The ratio of the size of an image to the size
of the object is called the magnification. It is given by
m = 1/
where y' is the height of the image and y is the height of the object. The second equality
allows you to find the size of the image (or object) with the information provided by the thin
lens equation.
All of the quantities in the above equations can take both positive and negative values.
Positive distances correspond to real images or objects, while negative distances
correspond to virtual images or objects. Positive heights correspond to upright images or
objects, while negative heights correspond to inverted images or objects. The following
table summarizes these properties:
positive negative
virtual
virtual
8 real
real
y upright
y upright
inverted
inverted
The focal length f can also be positive or negative. A positive focal length corresponds to a
converging lens, while a negative focal length corresponds to a diverging lens.
Consider an object with a = 12 cm that produces an image with 8 = 15 cm. Note that whenever you are working with a physical object, the object distance will be positive (in multiple optics setups, you will encounter
"objects" that are actually images, but that is not a possibility in this problem). A positive image distance means that the image is formed on the side of the lens from which the light emerges.
▾ Part A
Find the focal length of the lens that produces the image described in the problem introduction using the thin lens equation.
Express your answer in centimeters, as a fraction or to three significant figures.
f = 8.67 cm
Submit
✓ Correct
▾ Part B
Previous Answere
Considering the sign of f, is the lens converging or diverging?
converging
Odiverging
Submit
▾ Part C
Previous Answere
Correct
Review | Constants
Transcribed Image Text:Learning Goal: To learn the quantitative use of the lens equation, as well as how to determine qualitative properties of solutions. In working with lenses, there are three important quantities to consider. The object distance 8 is the distance along the axis of the lens to the object. The image distance s' is the distance along the axis of the lens to the image. The focal length f is an intrinsic property of the lens. These three quantities are related through the equation = + = = } Note that this equation is valid only for thin, spherical lenses. Unless otherwise specified, a lens problem always assumes that you are using thin, spherical lenses. The equation above allows you to calculate the locations of images and objects. Frequently, you will also be interested in the size of the image or object, particularly if you are considering a magnifying glass or microscope. The ratio of the size of an image to the size of the object is called the magnification. It is given by m = 1/ where y' is the height of the image and y is the height of the object. The second equality allows you to find the size of the image (or object) with the information provided by the thin lens equation. All of the quantities in the above equations can take both positive and negative values. Positive distances correspond to real images or objects, while negative distances correspond to virtual images or objects. Positive heights correspond to upright images or objects, while negative heights correspond to inverted images or objects. The following table summarizes these properties: positive negative virtual virtual 8 real real y upright y upright inverted inverted The focal length f can also be positive or negative. A positive focal length corresponds to a converging lens, while a negative focal length corresponds to a diverging lens. Consider an object with a = 12 cm that produces an image with 8 = 15 cm. Note that whenever you are working with a physical object, the object distance will be positive (in multiple optics setups, you will encounter "objects" that are actually images, but that is not a possibility in this problem). A positive image distance means that the image is formed on the side of the lens from which the light emerges. ▾ Part A Find the focal length of the lens that produces the image described in the problem introduction using the thin lens equation. Express your answer in centimeters, as a fraction or to three significant figures. f = 8.67 cm Submit ✓ Correct ▾ Part B Previous Answere Considering the sign of f, is the lens converging or diverging? converging Odiverging Submit ▾ Part C Previous Answere Correct Review | Constants
Part I
What is the x coordinate of the object? Keep in mind that a real image and a real object should be on opposite sides of the lens.
Express your answer in centimeters, as a fraction or to three significant figures.
z =
for Pärt for Pårt undo for Part redo for Part I reset for Part I keyboard shortcuts for Part I help for Part I
Submit Request Answer
cm
Transcribed Image Text:Part I What is the x coordinate of the object? Keep in mind that a real image and a real object should be on opposite sides of the lens. Express your answer in centimeters, as a fraction or to three significant figures. z = for Pärt for Pårt undo for Part redo for Part I reset for Part I keyboard shortcuts for Part I help for Part I Submit Request Answer cm
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