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A small candle 2.5 cm height is placed in front of a concave mirror of radius of curvature 36 cm at a distance 27 cm from the concave mirror. i.At what distance from the mirror should a screen be placed in order to receive a sharp image ? ii. Describe the nature and size of the image. iii.If the candle is moved close to the mirror, what will be the position of the screen ? |
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Answer» Solution :`h_(0) = 2.5 cm, R = 36 cm , f = (R)/(2) = ~ 18` cm , u = - 27 cm i. `(1)/(f) = (1)/(u) + (1)/(v) "" (1)/(v) = (1)/(f) - (1)/(u)""= (1)/(-18) - (1)/(-27)` v = - 54 cm The negative sign indicates that image is formed in front of mirror (on the same side as that of the object ) `therefore` screen should be placed at a distance of 54 cm in front of the mirror . ii.m= `(h_(i))/(h_(0)) = (-v)/(u) "" h_(i) = (-v)/(u) XX h_(0)= (-54)/(27 ) xx 2.5 = - 5 cm ` Image is magnified. Negative sign shows that image is real and inverted . iii.If the candle is moved CLOSER to the mirror, then u decreases and v increases. HENCE screen has to be moved farther and farther. When u `lt` 18 cm, image formed will be virtual and hence can.t be formed on the screen . |
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