A thin lens of focal length f and its apertur | vedclass

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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. Blocking a part of the aperture does no

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$f, 3I/4$

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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. Blocking a part of the aperture does not change these parameters, so the focal length remains $f$.The intensity $I$ of the image is proportional to the area of the aperture, which is $A = \pi (d/2)^2 = \pi d^2/4$.When the central part with diameter $d/2$ is blocked, the radius of the blocked part is $d/4$. The area of the blocked part is $A_{blocked} = \pi (d/4)^2 = \pi d^2/16$.The remaining area is $A_{remaining} = A - A_{blocked} = \frac{\pi d^2}{4} - \frac{\pi d^2}{16} = \frac{3\pi d^2}{16}$.The ratio of the remaining area to the original area is $\frac{3\pi d^2/16}{\pi d^2/4} = 3/4$.Therefore, the new intensity becomes $3I/4$.

$A$ thin lens of focal length $f$ and its aperture has a diameter $d$. It forms an image of intensity $I$. Now, the central part of the aperture up to diameter $(d/2)$ is blocked by an opaque paper. The focal length and image intensity would change to:

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