Question

A lens having a focal length $f$ and an aperture of diameter $d$ forms an image of intensity $I$. An aperture of diameter $\frac{d}{2}$ in the central region of lens is covered by a black paper. Focal length of the lens and intensity of the image now will be respectively:

• A. $f$ and $\frac{I}{4}$
• B. $\frac{3f}{4}$ and $\frac{I}{2}$
• C. $f$ and $\frac{3I}{4}$
• D. $\frac{f}{4}$ and $\frac{I}{2}$

Answer 1

The correct option is C $f$ and $\frac{3I}{4}$

The focal length of any spherical portion cut out from a sphere will be the same. It does not depend upon the shape of the spherical surface.

Intensity of the image is directly proportional to area of path through which light passes. On blocking the central part of lens, its focal length does not change. It remains $f$. Now, let initial area $A_1=\pi (\frac{d}{2}{)}^2=\frac{\pi d^2}{4}$

On blocking central part of aperture upto diameter $\frac{d}{2}$, let the area left out be $A_2$

$A_2=\pi (\frac{d}{2}{)}^2-\pi (\frac{d}{4}{)}^2$

$=\pi \frac{d^2}{4}-\pi \frac{d^2}{16}=\frac{3\pi d^2}{16}$

As, $\frac{I_2}{I_1}=\frac{A_2}{A_1}=\frac{3\pi d^2\times 4}{16\pi d^2}=\frac{12}{16}=\frac{3}{4}$

$I_2=\frac{3}{4}{I}_1=\frac{3}{4}{I}_1$