The distance between a screen and an object is 120 cm. A convex lens is placed closed to the object and is moved along the line joining object and screen, towards the screen. Two sharp images of the object are found on the screen. The ratio of magnification of two real images is 1: 9. Then,

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brightness of both the images is same

focal length of the lens is 22.5 cm

smaller image is brighter than the larger one

larger image is brighter than the smaller one

answer is A, B.

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Detailed Solution

Since object and image distances are exchangeable ${(\frac{v}{u})}^{2} = 9$
Using lens formula, we get f = 22.5 cm
In the second case lens is at greater distance from the object, it receives lesser part of light reaching it compared to when it is close to the lens in first case. As a result small image will be brighter than the large one.

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