In a photoelectric experiment,if both the intensity and frequency of the incident light are doubled,then the saturation photoelectric current

The number of photoelectrons emitted for light of a frequency $\nu$ (higher than the threshold frequency $\nu_0$) is proportional to

$A$ photosensitive material is at $9\,m$ to the left of the origin and the source of light is at $7\,m$ to the right of the origin along the $x$-axis. The photosensitive material and the source of light start from rest and move,respectively,with $8\widehat{i}\,m/s$ and $4\widehat{i}\,m/s$. The ratio of intensities at $t = 0$ to $t = 3\,s$ as received by the photosensitive material is:

Light of intensity $10^{-5} \; W m^{-2}$ falls on a sodium photo-cell of surface area $2 \; cm^2$. Assuming that the top $5$ layers of sodium absorb the incident energy,estimate the time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about $2 \; eV$. What is the implication of your answer?

The graph between the intensity of light $(I)$ falling on a metallic plate and the generated photoelectric current $(i)$ is:

Threshold wavelength for photoelectric emission from a metal surface is $5200\,\mathring{A}$. Photoelectrons will be emitted when this surface is illuminated with monochromatic radiation from: