an object is 1.0 cm tall and its erect image is 5.0 cm tall. what is the exact magnification?,

looking glass, magnification, glass @ Pixabay

What is the exact magnification of an object that is 1.0 cm tall and its erect image is 5.0 cm tall? The answer may surprise you! -An object that is magnified by a factor of M has an erect image with height = (M x H) ÷ (M + -H). The example we are using has the magnification 𝑛=0.040. The erect image in this case would be: V_{EI}={(0.040)(V_h)}/({(\ 041}{x}){+^{(-}\ \ {11})}}={\frac{{\left({40}{cm}\right)\times {\textwidth }}{m}}^{{-11}}=–|0|\\, where h is the original height and V_h is the original width. So if you know how tall something was, and how wide it was, you can find the exact magnification. You may have been expecting this to be a lot more complicated than that! But if you know an object’s original height and width, then there is only one formula for calculating its erect image magnification. As long as the two are in cm or mm units (or however your software calculates measurements), then they should convert automatically. The Erect Image Magnification by 𝑛=0.040 has V_{EI}={(0.040)(V_h)}/({(\ 041}{x}){+^{(-}\ \ {11})}}=-|-\\z{-|–|-||\textwidth } per unit of length


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