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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 the central pa

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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 the central part of the lens does not change these parameters,so the focal length remains $f' = f$.Intensity $I$ is proportional to the area of the aperture $A$. The initial area is $A = \pi (d/2)^2 = \pi d^2 / 4$.The area of the covered central part is $A_{covered} = \pi (d/4)^2 = \pi d^2 / 16$.The remaining area is $A' = A - A_{covered} = \frac{\pi d^2}{4} - \frac{\pi d^2}{16} = \frac{3\pi d^2}{16}$.Since $I \propto A$,the new intensity $I'$ is given by $I' = I 	imes (A'/A) = I 	imes (3/4) = 3I/4$.

$A$ 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,respectively?

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