Physics - Geometrical Optics Question with Solution | TestHub

The focal length remains$\mathrm{f}$, as any small portion of the lens forms a complete image of the object and has the same focal length.

While the intensity of the image depends on the area as,

$\mathrm{Intensity}\propto {\left(\mathrm{A}\mathrm{r}\mathrm{e}\mathrm{a}\right)}^2$

$\frac{{\mathrm{I}}_2}{{\mathrm{I}}_1}=\frac{{\mathrm{A}}_2}{{\mathrm{A}}_1}$

$\frac{{\mathrm{I}}_2}{\mathrm{I}}=\frac{\frac{\mathrm{\pi }{\mathrm{d}}^2}{4}-\frac{\mathrm{\pi }{\left(\frac{\mathrm{d}}{2}\right)}^2}{4}}{\frac{\mathrm{\pi }{\mathrm{d}}^2}{4}}$

$\frac{{\mathrm{I}}_2}{\mathrm{I}}=\frac{3}{4}$

${\mathrm{I}}_2=\frac{3}{4}\mathrm{I}$