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A concave spherical mirror can be used to project an image onto a sheet of paper, allowing the magnified image of an illuminated real object to be accurately traced. If you have a concave mirror with a focal length of 7.9 cm, where would you place a sheet of paper so that the image projected onto it is twice as far from the mirror as the object is? Is the image upright or inverted, real or virtual? What would the magnification be? location cm nature real virtual erect inverted magnification 

          A concave spherical mirror can be used to project an image onto a sheet of paper, allowing the magnified image of an illuminated real object to be accurately traced. If you have a concave mirror with a focal length of 7.9 cm, where would you place a sheet of paper so that the image projected onto it is twice as far from the mirror as the object is? Is the image upright or inverted, real or virtual? What would the magnification be?
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Added by Jessica K.

University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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A concave spherical mirror can be used to project an image onto a sheet of paper, allowing the magnified image of an illuminated real object to be accurately traced. If you have a concave mirror with a focal length of 7.9 cm, where would you place a sheet of paper so that the image projected onto it is twice as far from the mirror as the object is? Is the image upright or inverted, real or virtual? What would the magnification be? location cm nature real virtual erect inverted magnification
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The radius of curvature of a concave spherical mirror is 28.5 cm. (Make sure to use the appropriate sign in your final answers.) (a) Find the image distance and magnification if an object is placed 74.0 cm in front of the mirror? di = cm M = Is this image real or virtual? realvirtual Is it upright or inverted? upright inverted (b) Find the image distance and magnification if an object is placed 28.5 cm in front of the same mirror? di = cm M = Is this image real or virtual? real virtual Is it upright or inverted? upright inverted

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Transcript

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00:01 So from here we can say that one of the most important aspect of optics is the fact that the image can be projected if it must be real.
00:20 So from here if we assume that the object distance is equal to d0, then the image distance from here will be equal to 2 of d.
00:35 D node so we get that 1 divided by f is equal to 1 divided by the d not plus 1 divided by the 2 of d node so the question here is that the image distance ask about that image distance but it could be calculated by this formula so from here we can say that also the image is real so it has a positive value d mirror equation...
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