The distance between an object and its real image, formed by a converging lens, is held fixed. This distance is greater than four times the focal length of the lens. Show that there are two possible positions for the lens, and that the size of the object is given by sqrt(I_1 I_2) Where I_1 and I_2are the sizes of the two images.

The distance between an object and its real image, formed by a converg

1.	The distance between an object and its real image, formed by a converging lens, is held fixed. This distance is greater than four times the focal length of the lens. Show that there are two possible positions for the lens, and that the size of the object is given by sqrt(I_1 I_2) Where I_1 and I_2are the sizes of the two images.
Answer» O=I1/I2 O=I1I2 O=\\| I1I2 O=I1-I2 Solution :`m_1 = (I_1)/(O) = v/u "" ...(1)` Diminished image of size `I_2` for the POSITION `L_2` of the lens ` m_2 = (I_2)/(O) = v/u`....(2) From (1) and (2) , ` (I_1 I_2)/(O^2) = v/u XX u/v = 1` ` O = sqrt(I_1 I_2)`

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