This question will look at the intensity of radiation in a bit more detail. The intensity of light, $I$, incident normally on an area $A$ is defined to be $I=\frac{p}{A}$, where $P$ is the power carried by the light. Show that $I=\varphi hf$, where $\varphi$ is the photon flux density, i.e, the number of photons incident on the surface per second per unit area, and $f$ is the frequency of the light.

Calculate the intensity of light of wavelength $\lambda =5.0\times {10}^{-7} \mathrm{m}$ Incident on a surface when the photon flux density is $\varphi =3.8\times {10}^{18} {\mathrm{m}}^{-2}{\mathrm{s}}^{-1}.$

The wavelength of the light is decreased to $\lambda =4.0\times {10}^{-7}\mathrm{m}.$ Calculate the new photon flux density so that the intensity of light incident on the surface is the same as that found in $\mathbf{b}$.

Note-In case of $\mathbf{b}$ the intensity of light$=1.5 {\mathrm{Wm}}^{-2}$.

Explain why light of wavelength $\lambda =4.0\times {10}^{-7} \mathrm{m}$ and of the same intensity as that of light of wavelength $\lambda =5.0\times {10}^{-7}\mathrm{m}$ will result in fewer electrons being emitted from the surface emitted from the surface per second.

Hence explain why light of wavelength $\lambda =4.0\times {10}^{-7} \mathrm{m}$ and of the same intensity as that of light of wavelength $\lambda =5.0\times {10}^{-7}\mathrm{m}$ will result in fewer electrons being emitted from the surface emitted from the surface per second. State one assumption made in reaching to this conclusion.