1.

A lens having focal length f and aperture of diameter d forms an image of intensity I. Aperture of diameter (d)/(2) is central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively.

Answer»

`f and (3I)/(4)`
`(f)/(2) and (I)/(2)`
`f and (I)/(4)`
`2f`

Solution :When a part of LENS aperture is covered by a black paper, the focal length of lens remains UNCHANGED. However, when central portion of diameter `(d)/(2)` is covered out of diameter d, then area `A.=(1)/(4)pi((d)/(2))^(2)=(1)/(4)A` is blackened and light rays can PAS through remaining area `A-A.=(3)/(4)A`. Hence, INTENSITY of final image is `(2I)/(4)`.


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