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An object 1.5 cm is size is placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is 40 cm. Detemine the maguificatin produced by the two-lens system an the size of the image. |
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Answer» Solution : `(1)/(v_(1)) - (1)/(u_(1)) = (1)/(f_(1)) "" (1)/(v_(1)) - (1)/((-40))= (1)/(30) ` `(1)/(v_(1)) = (1)/(30) - (1)/(40) = (1)/(120) "" THEREFORE v_(1) = 120 cm ` `m_(1) = (v_(1))/(|u_(1)|)= (120)/(40)= 3` `u_(2) = 120 - 8 = 112 cm , f_(2)=- 20 ` cm `(1)/(v_(2)) - (1)/(u_(2)) = (1)/(f_(2)) "" (1)/(v_(2)) - (1)/(112)= (1)/(-20)` `(1)/(v_(2)) = (1)/(112) - (1)/(20) = (-23)/(560)"" therefore v_(2) = (-560)/(23) cm "" m_(2) =(|v_(2)|)/(u_(2)) = (((560)/(23)))/(122)= (5)/(23)` Net MAGNIFICATION, m=`m_(1) m_(2) = (3xx5)/(23) APPROX 0.65 ` Size of the image = 0.98 cm |
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