Question

A thin lens of focal length $f$ and aperture diameter $d$ forms an image of intensity $I$. If the central part of the aperture upto diameter $d/2$ is blocked by an opaque paper, then the new focal length and intensity of image will be

• A. $\frac{f}{2},\frac{l}{2}$
• B. $\frac{f}{2},\frac{3}{4}l$
• C. $f,\frac{1}{2}$
• D. $f,\frac{3}{4}l$

Answer 1

The correct option is D $f,\frac{3}{4}l$

On blocking the central part of the lens, focal length does not change as $f$ remains same.
Intensity of image is directly proportional to the area of lens.
Initial area, $A_1=\pi {\left(\frac{d}{2}\right)}^2=\frac{\pi d^2}{4}$
On blocking, the central part of the aperture upto diameter $\frac{d}{2}$, the new area
$A_2=\pi {\left(\frac{d}{2}\right)}^2-\pi {\left(\frac{d}{4}\right)}^2=\frac{\pi d^2}{4}-\frac{\pi d^2}{16}$
$=\frac{3\pi d^2}{16}$
As $\frac{l_2}{l_1}=\frac{A_2}{A_1}=\frac{3\pi d^2\cdot 4}{16\pi d^2}=\frac{12}{16}=\frac{3}{4}$
$\therefore l_1=\frac{3}{4}{l}_1$.