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An object is placed 30 cm away from a symmetric convex lens and an image two thirds of the size of the object is produced. The object is moved by a distance of 20 cm so as to get a magnified image. Now we get (a) a real image of magnification (b) a virtual image of magnification 5 (c) a real image at a distance of 40 cm (d) a virtual image at a distance of 60 cm |
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Answer» Correct option (d) a virtual image at a distance of 60 cm Explanation: Size of the image : size of the object ⇒ v = 2/3 u Lens formula (real image) : 1/v - 1/u = 1/f ⇒ 3/2u + 1/u = 1/f ⇒ f = 2u/5 = 2 x 30 /5 = 12 cm After moving the object by 20 cm, it is with in the focus of the lens. So it forms a virtual image on the same side as that of the object. Lens formula (virtual image : 1/v - 1/u = 1/f ⇒ - 1/v + 1/10 = 1/12 ⇒ 1/v = 1/60 ⇒ v = 60 cm |
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