Physics - Wave Optics Question with Solution | TestHub

$\mathrm{I}=\frac{\text{P}\text{of source}}{4\mathrm{\pi }{\left(\text{distance}\right)}^2}=\frac{\mathrm{P}}{4{\mathrm{\pi d}}^2}$
Here, we assume light to spread uniformly in all directions.
The number of photo-electrons emitted from a surface depend on the intensity of light I falling on it. Thus the number of electrons emitted n depends directly on I. P remains constant as the source is the same.

$\therefore$$\genfrac{}{}{0.1ex}{}{{\mathrm{I}}_2}{{\mathrm{I}}_1}=\genfrac{}{}{0.1ex}{}{{\mathrm{n}}_2}{{\mathrm{n}}_1}\Rightarrow \genfrac{}{}{0.1ex}{}{{\mathrm{P}}_2}{{\mathrm{P}}_1}{\left(\genfrac{}{}{0.1ex}{}{{\mathrm{d}}_1}{{\mathrm{d}}_2}\right)}^2=\genfrac{}{}{0.1ex}{}{{\mathrm{n}}_2}{{\mathrm{n}}_1}$

$\therefore$$\genfrac{}{}{0.1ex}{}{{\mathrm{n}}_2}{{\mathrm{n}}_1}=\left(\genfrac{}{}{0.1ex}{}{\mathrm{P}}{\mathrm{P}}\right){\left(\genfrac{}{}{0.1ex}{}{1}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right)}^2=\genfrac{}{}{0.1ex}{}{4}{1}$$(\because \mathrm{P}1 = \mathrm{P}2 = \mathrm{P})$