A screen is kept at a distance of 1 m from the object. A converging lens when placed between the object and screen, at any two positions 80 cm apart, forms a sharp image of the object on the screen. If the size of the object is 6 mm, then

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The size of image at second position when lens is nearer to screen than object is 54 mm

The magnitude of magnification at first position when object is near to the lens is 9

The product of magnification m1 and m2 at two position is –1

Focal length of lens is 9 cm

answer is A, B.

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Detailed Solution

$f = \frac{D^{2} - d^{2}}{4 D}$

$= \frac{{(100)}^{2} - {(80)}^{2}}{400}$

$f = \frac{10000 - 6400}{400} = 9 c m$

$m_{1} = \frac{D + d}{D - d} = \frac{100 + 80}{100 - 80} = \frac{180}{20} = 9$

$m_{1} . m_{2} = 1$

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