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:

Updated On: Nov 26, 2024
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The Correct Option is C

Solution and Explanation

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.

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