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Constants Consider an object with s = 12 cm that produces an image with s' = 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. Learning Goal: In working with lenses, there are three important quantities to consider: The object distance s 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 Part A 1+1= + = s f 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. 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. VAD ? 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 f = cm Submit Request Answer & m = y' y — — 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. Part B Complete previous part(s) Part C Complete previous part(s) Part D Complete previous part(s) 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 Now consider a diverging lens with focal length f= -15 cm, producing an upright image that is 5/9 as tall as the object. =
Constants What 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 s. Express your answer in centimeters, as a fraction or to three significant figures. Learning Goal: In working with lenses, there are three important quantities to consider: The object distance s 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 ΤΙ ΑΣΦ ? 1+ } = } S = S cm = 1 f s 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. Submit Request Answer Part G Complete previous part(s) 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 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. = m = m y' y = S Part H 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. What is the image distance? Express your answer in centimeters, as a fraction or to three significant figures. 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: ΤΕΙ ΑΣφ ? nositive negative cm