question_answer

A source of light is placed at a distance\[\mathbf{50}\text{ }\mathbf{cm}\] from a photo-cell and the cut-off potential is found to be \[{{V}_{0}}\]. If the distance between the light source and photo-cell is made\[\mathbf{25}\text{ }\mathbf{cm}\], what will be the new cut-off potential? Justify your answer.

Answer:

                The stopping potential is still \[{{V}_{0}}\]. As the distance is decreased from \[\text{5}0\text{ cm}\] to \[\text{25 cm}\] (i.e., the distance is halved), the intensity of light becomes four times the original intensity. But the stopping potential is independent of the intensity.