a convergiing lens produces a real ,magnified and well defined image of a small illuminated square on a screen .the area of the image A1. when the lens is moved towards the screen without disturbing the object and the screen ,the area of the well defined image obtained on the screen is A2.what is the side of the squre object.
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Explanation:
The two positions (of the lens) for which well defined images of the square are obtained, are the conjugate positions and hence we have
a =√(a1a2) ,
where a1 and a2 are the sides of the images in the two cases and a is the side of the square object.
[The linear magnification in the first case is v/u = a1/a.
Since the object distance u and the image distance v are interchanged in conjugate positions, we have, in the second case,
u/v = a2/a.
From the above expressions, 1 = a1a2/a2 from which a =√(a1a2)].
But a1 = √A1 and a2 = √A2
Therefore, a =√(√A1√A2) = (A1A2)^1/4
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