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An object of height 2.5 cm is placed perpendicularly on the principal axis of a concave mirror of focal length f at a distance of (3)/(4)f. What will be the nature of the image of the object and its height? |
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Answer» Solution :For concave mirror, the FOCAL length `f` is actually - f, using the proper sign. Also, u = `-(3)/(4)f.` So the relation `(1)/(v) + (1)/(u) = (1)/(f)` gives, `(1)/(v) + (-(4)/(3f)) = (1)/(-f) or, (1)/(v) = (4)/(3f) - (1)/(f) = (1)/(3f)` or, v = 3f The positive sign of v means that the image is formed on the OPPOSITE side, so it is a virtual image. MAGNIFICATION = `-(v)/(u) = (3f)/(-(3f//4)) = 4 ` `therefore` Image height `= 4 xx 2.5 = 10 `cm |
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