Question

Which figure shows the correct variation of applied potential difference ($V$) with photoelectric current ($I$) at two different intensities of light ($I_1 < I_2$) of same wavelengths:

• A.
• B.
• C.
• D.

Answer 1

Answer: (C)

Answer 2

Understanding the Photoelectric Effect and Current-Voltage Characteristics:
In the photoelectric effect, the photoelectric current $I$ depends on the intensity of the incident light and the applied potential difference $V$.

The key points to note are:
- For a given frequency (or wavelength), the stopping potential $V_0$ remains constant, as it depends only on the frequency of the incident light and not on its intensity.
- The saturation current (maximum current) is proportional to the intensity of the incident light. Therefore, higher intensity light (e.g., $I_2$) results in a higher saturation current than lower intensity light (e.g., $I_1$).

Selecting the Correct Graph:
Since the wavelength (or frequency) of the light is the same for both intensities, the stopping potential $V_0$ will be the same for both $I_1$ and $I_2$. However, since $I_2 > I_1$, the saturation current for $I_2$ will be greater than that for $I_1$.
Option (3) correctly shows:
- The stopping potential $V_0$ is the same for both intensities.
- The saturation current for $I_2$ is greater than for $I_1$, consistent with the higher intensity of $I_2$.

Conclusion:
Therefore, the correct graph is Option (3) , as it accurately represents the photoelectric current variation with applied potential for two different light intensities of the same wavelength.