the only parts that i need help with is parts g and h + 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 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 by=//mwhere 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 negativevirtual8 realrealy uprighty uprightvirtualinvertedinvertedThe 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.▾Consider an object with 8 = 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 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 AFind 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▾ Part B✓ CorrectConsidering the sign of f, is the lens converging or diverging?convergingOdivergingSubmitPrevious Answers▾ Part C✓ CorrectSubmitPrevious AnswereWhat is the magnification m of the lens?Express your answer as a fraction or to three significant figures.7 -1.25Part DPrevious Answers✔Correct< 10 of 22Review | Constants 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.mThe 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 byVwhere 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:8 reapositive negativevirtualvirtualrealy uprighty uprightinvertedinvertedThe 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 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.8 = 12.0 cmSubmitAll attempts used; correct answer displayedPart GWhat is the image distance?Express your answer in centimeters, as a fraction or to three significant figures.SubmitPrevious Answerefor Part for Part Ado for Part&redo fort Gresor Part G keyboard shortcuts for Part G help for Part Ga vonPart HX Incorrect; Try Again; 5 attempts remaining=1512Previous Answers Request AnswerA lens placed at the origin with its axis pointing along the x axis produces a real inverted image at z = -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.Submitor Part Hfor avec1512do for Part redo forart H restor Part H keyboard shortcuts for Part H help for Part HcmPrevious Anawara Requent Anawarcm

Answered: Image /qna-images/answer/284ec1c1-315e-46ab-a75f-abb08b23651c.jpg

a ndly. ✓ in o a Part G What is the image distance? Express your answer in centimeters, as a fraction or to three significant figures. for Part for Part Ado for Part&redo fort G resor Part G keyboard shortcuts for Part G help for Part G vec Submit Previous Answers Request Answer 15 12 * Incorrect; Try Again; 5 attempts remaining Part H A lens placed at the origin with its axis pointing along the x axis produces a real inverted image at = -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. for Part vec 15 12 Submit Mo for Part redo foart H resor Part H keyboard shortcuts for Part H help for Part H cm Previous Anawara Request Anawar cm

a ndly. ✓ in o a Part G What is the image distance? Express your answer in centimeters, as a fraction or to three significant figures. for Part for Part Ado for Part&redo fort G resor Part G keyboard shortcuts for Part G help for Part G vec Submit Previous Answers Request Answer 15 12 * Incorrect; Try Again; 5 attempts remaining Part H A lens placed at the origin with its axis pointing along the x axis produces a real inverted image at = -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. for Part vec 15 12 Submit Mo for Part redo foart H resor Part H keyboard shortcuts for Part H help for Part H cm Previous Anawara Request Anawar cm

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter23: Mirrors And Lenses

Section: Chapter Questions

Problem 8P: To fit a contact lens to a patient's eye, a keratometer can be used to measure the curvature of the...

See similar textbooks

Similar questions

A certain slide projector has a 100 mm-focal length lens. (a) How far away is the screen if a slide is placed 103 mm from the lens and produces a sharp image? (b) If the slide is 24.0 by 36.0 mm, what are the dimensions of the image? Explicitly show how you follow the steps in the Problem-solving strategy: Lenses.
Andy decides to use an old pair of eyeglasses to make some optical instruments. He knows that the near point in his left eye is 50.0 cm and the near point in his right eve is 100 cm. (a) What is the maximum angular magnification he can produce in a telescope? (b) If he places the lenses 10.0 cm apart, what is the maximum overall magnification he can produce in a microscope? Hint: Go back to basics and use the thin lens equation to solve part (b).
A lens is used to examine an object across a room. Is the lens probably being used as a simple magnifier? Explain in terms of focal length, the image, and magnification.
A person with a nearsighted eye has near and far points of 16 cm and 25 cm, respectively. (a) Assuming a lens is placed 2.0 cm from the eye, what power must the lens have to correct this condition? (b) Suppose contact lenses placed directly on the cornea are used to correct the person's eyesight. What is the power of the lens required in this case, and what is the new near point? Hint: The contact lens and the eyeglass lens require slightly different powers because they are at different distances from the eye.
A certain slide projector has a 100 mm focal length lens. (a) How far away is the screen, if a slide is placed 103 mm from the lens and produces a sharp image? (b) If the slide is 24.0 by 36.0 mm, what are the dimensions of the image? Explicitly show how you follow the steps in the PreblemSolving Strategy for lenses.
A person with a nearsighted eye has near and far points of 16 cm and 25 cm, respectively. (a) Assuming a lens is placed 2.0 cm from the eye, what power must the lens have to correct this condition? (b) Suppose contact lenses placed directly on the cornea are used to correct the person's eyesight. What is the power of the lens required in this case, and what is the new near point? Hint: The contact lens and the eyeglass lens require slightly different powers because they are at different distances from the eye.
A diverging lens has a focal length of 20.0 cm. Use graph paper to construct accurate ray diagrams for object distance of (a) 40.0cm and (b) 10.0 cm. In each case, determine the location of the image from the diagram and the image magnification, and state whether the image is uptight or inverted. (c) Estimate the magnitude of uncertainty in locating the points in the graph. Are your answers and the uncertainty consistent with the algebraic answers found in Problem 33?
The shell” of a concept map dealing with lenses and their properties is shown in Concept Map 9.2. Most of the concepts and all of the linking phrases needed to form meaningful propositions have been left out. Complete this map by selecting the appropriate concepts and linkages from the lists that follow. Some connecting words will have to be used more than once to finish the map correctly. If you need to, be sure to review Sections 9.3 through 9.5 for help in properly determining and relating the relevant concepts.
Objects viewed through a microscope are placed very close to the focal point of the objective lens. Show that the minimum separation x of two objects resolvable through the microscope is given by x=1.22f0D , where f0is the focal length and D is the diameter of the objective lens as shown below.
People who do very detailed work close up, such as jewellers, often can see objects clearly at much closer distance than the normal 25 cm. (a) What is the power of the eyes of a woman who can see an object clearly at a distance of only 8.00 cm? (b) What is 1J1e size of an image of a 1.00 mm object, such as lettering inside a ring, held at this distance? (c) What would the size of the image be if the object were held at the normal 25.0 cm distance?
(a) Use the thin-lens equation to derive an expression for q in terms of f and p. (b) Prove that for a real object and a diverging lens, the image must always be virtual. Hint: Set f = |f| and show that q must be less than zero under the given conditions. (c) For a real object and converging lens, what inequality involving p and f must hold if the image is to be real?
Suppose a 200 mm focal length telephoto lens is being used to photograph mountains 10.0 km away. (a) Where is file image? (b) What is the height of the image of a 1000 m high cliff on one of the mountains?