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What should be the distance between the object in Exercise 28 and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm^(2) Would you be able to see the squares distinctly with your eyes very close to the magnifier? |
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Answer» Solution :New area = 6.25 `mm^(2)` `THEREFORE` LENGTH of each side = 2.5 mm Linear magnification = `(v)/(U) = (2.5)/(1) = 2.5` v = 2.5 u, `"" therefore (1)/(v) - (1)/(u) = (1)/(f) ` `(1)/(v)- (1)/(u) = (1)/(2.5 u) - (1)/(u) = (1 - 2.5)/(2.5u) = (1.5)/(2.5 u) = (1)/(10)` i.e., u = - 6 CM since |v| = 15 cm `lt` 25 cm. The imagelies at a distance LESS than the least distance of distance vision and cannot be observed distanctly. |
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