A lens of focal length f and diameter d forms | vedclass

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(D) The focal length $f$ of a lens depends only on the radii of curvature of its surfaces and the refractive index of the material. It is independent

Vedclass · Accepted Answer

$f$ and $\frac{3I}{4}$

Vedclass · Answer

(D) The focal length $f$ of a lens depends only on the radii of curvature of its surfaces and the refractive index of the material. It is independent of the aperture (diameter) of the lens. Therefore,the focal length remains $f$.The intensity $I$ of the image formed by a lens is directly proportional to the area of the aperture,$I \propto A$.Initially,the area $A = \pi (\frac{d}{2})^2 = \frac{\pi d^2}{4}$.When the central part of diameter $\frac{d}{2}$ is covered,the effective area $A'$ is the total area minus the area of the covered part:$A' = \pi (\frac{d}{2})^2 - \pi (\frac{d/2}{2})^2 = \frac{\pi d^2}{4} - \frac{\pi d^2}{16} = \frac{3\pi d^2}{16}$.The ratio of the new intensity $I'$ to the original intensity $I$ is:$\frac{I'}{I} = \frac{A'}{A} = \frac{3\pi d^2 / 16}{\pi d^2 / 4} = \frac{3}{4}$.Thus,the new intensity is $I' = \frac{3I}{4}$.

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

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