Physics, asked by pushkarburnwalpuru, 1 day ago

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Answered by tagorbisen
2

Explanation:

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When the object is located at a location beyond the center of curvature, the image will always be located somewhere in between the center of curvature and the focal point. Regardless of exactly where the object is located, the image will be located in the specified region. In this case, the image will be an inverted image. That is to say, if the object is right side up, then the image is upside down. In this case, the image is reduced in size; in other words, the image dimensions are smaller than the object dimensions. If the object is a six-foot tall person, then the image is less than six feet tall. The magnification is the ratio of the height of the image to the height of the object. In this case, the absolute value of the magnification is less than 1. Finally, the image is a real image. Light rays actually converge at the image location.

Answered by souhardya51
0

Answer:

When the object is located at a location beyond the center of curvature, the image will always be located somewhere in between the center of curvature and the focal point. Regardless of exactly where the object is located, the image will be located in the specified region. In this case, the image will be an inverted image. That is to say, if the object is right side up, then the image is upside down. In this case, the image is reduced in size; in other words, the image dimensions are smaller than the object dimensions. If the object is a six-foot tall person, then the image is less than six feet tall. The magnification is the ratio of the height of the image to the height of the object. In this case, the absolute value of the magnification is less than 1. Finally, the image is a real image. Light rays actually converge at the image location.

Explanation:

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