Learning Goal:To learn the quantitative use of the lens equation, as well as how to determine qualitativeproperties of solutions.In working with lenses, there are three important quantities to consider. The object distances is the distance along the axis of the lens to the object. The image distance s' is thedistance along the axis of the lens to the image. The focal length f is an intrinsic propertyof the lens. These three quantities are related through the equation==+=+Note that this equation is valid only for thin, spherical lenses. Unless otherwise specified, alens 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 areconsidering a magnifying glass or microscope. The ratio of the size of an image to the sizeof the object is called the magnification. It is given by흠.where y' is the height of the image and y is the height of the object. The second equalityallows you to find the size of the image (or object) with the information provided by the thinlens equation.m=1/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 distancescorrespond to virtual images or objects. Positive heights correspond to upright images orobjects, while negative heights correspond to inverted images or objects. The followingtable summarizes these properties:positive negativevirtualvirtualy upright invertedy upright inverted8 realrealThe focal length ƒ can also be positive or negative. A positive focal length corresponds to aconverging lens, while a negative focal length corresponds to a diverging lens.Submit✓ Correct▾ Part FWhat is the object distance? You will need to use the magnification equation to find a relationship between s and s'. Then substitute into the thin lens equation to solve for 8.Express your answer in centimeters, as a fraction or to three significant figures.for Parke For Eart do for Part redo foart Frelor Part F keyboard shortcuts for Fart F help for Part FSubmitPrevious Answers▾ Part HRequest AnswerPart G Complete previous part(s)A lens placed at the origin with its axis pointing along the x axis produces a real inverted image at x = -24 cm that is twice as tall as the object.What is the image distance?Express your answer in centimeters, as a fraction or to three significant figures.Submitfor Part for Part Ho for Part redo forent H restor Part H keyboard shortcuts for Part H help for Part HvecRequest AnswerPart I Complete previous part(s)Part J Complete previous part(s)cmPart K Complete previous part(s)cm + Understanding LensesLearning Goal:To learn the quantitative use of the lens equation, as well as how to determine qualitativeproperties of solutions.In working with lenses, there are three important quantities to consider. The object distance8 is the distance along the axis of the lens to the object. The image distance s' is thedistance along the axis of the lens to the image. The focal length f is an intrinsic propertyof the lens. These three quantities are related through the equation= + ===Note that this equation is valid only for thin, spherical lenses. Unless otherwise specified, alens 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 areconsidering a magnifying glass or microscope. The ratio of the size of an image to the sizeof the object is called the magnification. It is given by4Vm =where y' is the height of the image and y is the height of the object. The second equalityallows you to find the size of the image (or object) with the information provided by the thinlens 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 distancescorrespond to virtual images or objects. Positive heights correspond to upright images orobjects, while negative heights correspond to inverted images or objects. The followingtable summarizes these properties:positive negative8 realvirtualrealvirtualy uprightinvertedy' upright invertedThe focal length f can also be positive or negative. A positive focal length corresponds to aconverging lens, while a negative focal length corresponds to a diverging lens.▾ Part AConsider 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.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.Painfo Far do for Part redo foart A reor Part A keyboard shortcuts for Part A help for Part Af=SubmitPart B Complete previous part(s)Part C Complete previous part(s)Part D Complete previous part(s)Now consider a diverging lens with focal length f= -15 cm, producing an upright image that is 5/9 as tall as the object.▾ Part EIs the image real or virtual? Think about the magnification and how it relates to the sign of s'realRequest AnswervirtualSubmit▾ Part FPrevious Answere✔Correct.cmWhat is the object distance? You will need to use the magnification equation to find a relationship between s and s'. Then substitute into the thin lens equation to solve for 8.Express your answer in centimeters, as a fraction or to three significant figures.Pearson10 of 22Review | Constant

Since you have posted a question with multiple sub-parts, we will solve the first three sub-parts…

Consider an object with = 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 encou "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. Panor Part Vdo for Panredo foar A reor Part A keyboard shortcuts for Part A help for Part A f= Submit Part B Complete previous part(s) Part C Complete previous part(s) Part D Complete previous part(s) Now consider a diverging lens with focal length f-15 cm, producing an upright image that is 5/9 as tall as the object. Part E Request Answer Is the image real or virtual? Think about the magnification and how it relates to the sign of s @real virtual ▾ Part F Previous Answers Correct cm What is the object distance? You will need to use the magnification equation to find a relationship between aand s'. Then substitute into the thin lens equation to solve for a Express your answer in centimeters, as a fraction or to three significant figures.

11th Edition

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)...