# An object is placed in front of a screen and a convex lens is placed at a position such that the size of the image formed is 9cm.

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An object is placed in front of a screen and a convex lens is placed at a position such that the size of the image formed is 9cm. when the lens is shifted through a distance of 20 cm, the size of the image becomes 1 cm. The focal length of the lens and the size of the object are respectively.

A. 7.5 cm and 3.5 cm B. 7.5 cm and 4.5 cm

C. 6 cm and 3 cm D. 7.5 cm and 3 cm

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If h1 and h2 are the sizes of the images formed in two conjugate positions, then, the size of object is given as h = √h1h2 = √9×1 = 3cm.

The size of the image in the two conjugate positions is

In the first case, if the image is formed, the magnification is given as

In the first case, if the image is formed, the magnification is given as $\frac{v}{u} = \frac{9}{3}$Such that, v = 3u

Also, v = 20+u, since “v” and “u” interchange the conjugate positions.

∴ 3u = 20 + u

Or 2u = 20

Or u = 10

And, v = 20+ u = 20+10 = 30 cm

Now, focal length is calculated as

$\frac1f=\frac1v+\frac1u$

$\frac{1}{30} + \frac{1}{10}$

$\frac{10+30}{300}$

$\frac{40}{300} = 7.5$

Thus, correct answer is option (d)