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A lens having focal length f and aperture of diameter d forms an image of intensity I. Aperture of diameter $\frac{d}{2}$ in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively
A
$\frac{f}{2}$ and $\frac{I}{2}$
B
f and $\frac{I}{4}$
C
$\frac{3f}{4}$ and $\frac{I}{2}$
D
f and $\frac{3I}{4}$
Detailed Solution
The focal length depends only on the refractive index and the radii of curvature of the lens, not on its aperture; so it remains f.
The intensity of the image is proportional to the area of the lens that transmits light, i.e., $I \propto d^2$.
Original transmitting area $\propto d^2$; area covered $\propto \left(\frac{d}{2}\right)^2 = \frac{d^2}{4}$
Remaining area $\propto d^2 - \frac{d^2}{4} = \frac{3d^2}{4}$
New intensity $= I - \frac{I}{4} = \frac{3I}{4}$
So the focal length is f and the intensity is $\frac{3I}{4}$.
The intensity of the image is proportional to the area of the lens that transmits light, i.e., $I \propto d^2$.
Original transmitting area $\propto d^2$; area covered $\propto \left(\frac{d}{2}\right)^2 = \frac{d^2}{4}$
Remaining area $\propto d^2 - \frac{d^2}{4} = \frac{3d^2}{4}$
New intensity $= I - \frac{I}{4} = \frac{3I}{4}$
So the focal length is f and the intensity is $\frac{3I}{4}$.