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(D) The focal length $f$ of a lens depends on its refractive index and the radii of curvature of its surfaces. Covering a part of the lens does not ch

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$f$ and $3I/4$

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(D) The focal length $f$ of a lens depends on its refractive index and the radii of curvature of its surfaces. Covering a part of the lens does not change these properties,so the focal length remains $f$.The intensity $I$ of the image formed by a lens is directly proportional to the area $A$ of the aperture through which light passes $(I \propto A)$.The initial area of the aperture is $A_1 = \pi (d/2)^2 = \pi d^2/4$.When the central region with diameter $d/2$ (radius $d/4$) is covered,the area of the covered part is $A_{covered} = \pi (d/4)^2 = \pi d^2/16$.The new area of the aperture through which light passes is $A_2 = A_1 - A_{covered} = \frac{\pi d^2}{4} - \frac{\pi d^2}{16} = \frac{3\pi d^2}{16}$.The ratio of the new intensity $I_2$ to the initial intensity $I_1$ is given by the ratio of the areas: $\frac{I_2}{I_1} = \frac{A_2}{A_1} = \frac{3\pi d^2/16}{\pi d^2/4} = \frac{3}{4}$.Therefore,the new intensity is $I_2 = \frac{3}{4}I$ and the focal length remains $f$.

$A$ lens having focal length $f$ and aperture of diameter $d$ forms an image of intensity $I$. If the central region of the lens with aperture 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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