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

f and 1/4

3f/4 and 1/2

f and 3I/4

f/2 and I/2

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Solution

f and 3I/4

Intensity space space straight I proportional to straight A squared rightwards double arrow space straight I subscript 2 over straight I subscript 1 space equals space open square brackets straight A subscript 2 over straight A subscript 1 close square brackets squared space equals space fraction numerator πr squared space minus begin display style πr squared over 4 end style over denominator πr squared end fraction space equals space 3 divided by 4 rightwards double arrow space straight I subscript 2 space equals space 3 over 4 straight I subscript 1 space and space focal space length space remains space unchaged.

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