A convex lens forms an image of an object on a screen. The height of the image is 9 cm. The lens is now displaced until an image is again obtained on the screen. The height of this image is 4 cm. The distance between the object and the screen is 90 cm.

The distance between the two positions of the lens is 30 cm.

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The distance of the object from the lens in its first position is 36 cm.

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The height of the object is 6 cm.

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The focal length of the lens is 21.6 cm.

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Solution

The correct option is D The focal length of the lens is 21.6 cm.
$h_{o b j e c t}^{2} = h_{i m a g e 1} \times h_{i m a g e 2}$

$h_{o b j e c t} = \sqrt{36} = 6$

magnification of image is $\frac{v}{u} = \frac{9}{6}$

$v = \frac{3 u}{2}$

in lens displacement method , $u + v = d$ ; $u v = d f$

$u + \frac{3 u}{2} = 90$

$u = 36$ => $v = 54$

$u v = d f$

$f = \frac{36 \times 54}{90} = 21.6$

option $D$ is correct

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A convex lens forms an image of an object on a screen. The height of the image is 9 cm. The lens is now displaced until an image is again obtained on the screen. The height of this image is 4 cm. The distance between the object and the screen is 90cm.

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