A lens having focal length f and aperture of | vedclass

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(D) The focal length $f$ of a lens depends on the refractive index of the material and the radii of curvature of its surfaces. Covering a part of the

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$f$ and $3I/4$

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(D) The focal length $f$ of a lens depends on the refractive index of the material and the radii of curvature of its surfaces. Covering a part of the lens does not change these parameters,so the focal length remains $f$.Intensity $I$ is proportional to the area of the aperture $A = \pi (d/2)^2 = \pi d^2/4$.When the central part with diameter $d/2$ is covered,the radius of the covered part is $d/4$. The area of the covered part is $A_{covered} = \pi (d/4)^2 = \pi d^2/16 = A/4$.The remaining area of the lens is $A' = A - A/4 = 3A/4$.Since intensity $I$ is proportional to the area,the new intensity $I'$ is $I' = I 	imes (A'/A) = I 	imes (3/4) = 3I/4$.

$A$ lens having focal length $f$ and aperture of diameter $d$ forms an image of intensity $I$. If the central region of the lens with diameter $\frac{d}{2}$ is covered by black paper,what will be the new focal length and intensity of the image?

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