Animated Solution for Physics - Dual Nature of Matter and Radiation: The figure shows the variation of photocurrent with anode potential for a photosensitive surface for three different radiations. Let $I_a,I_b$ and $I_c$ be the intensities and $f_a,f_b$ and $f_c$ be the frequencies for the curves $a,b$ and $c$ respectively. Then :

Imagine you are conducting a photoelectric experiment. You shine light on a metal surface and measure the resulting photocurrent as you vary the anode potential. The graph in front of us is the visual diary of this exact experiment, performed with three different light radiations: $a$, $b$, and $c$.

The y-axis represents the photocurrent ($I$), which tells us how many electrons are reaching the anode per second. The x-axis represents the anode potential ($V$), which is the voltage we apply to either accelerate or stop these electrons.

Let's look at the left side of the graph, where the voltage is negative. This is the retarding potential. The exact point where the photocurrent drops to zero is called the stopping potential ($V_0$).

According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted electrons depends strictly on the frequency of the incident light: Since $e{V}_0=K_{\max }$, the stopping potential is a direct indicator of the light's frequency.

Notice how curves $a$ and $b$ merge and hit the x-axis at the exact same point? This means they share the same stopping potential. Therefore, the frequency of radiation $a$ must be equal to the frequency of radiation $b$: Curve $c$, on the other hand, has a more negative stopping potential, meaning its frequency $f_c$ is higher than both $f_a$ and $f_b$.

Now, let's shift our focus to the right side of the graph, where the voltage is positive. Notice how the curves eventually flatten out? This flat region is the saturation current ($I_s$). At this point, every single electron emitted from the cathode is being successfully collected by the anode.

The saturation current is directly proportional to the intensity of the incident light. More intensity means more photons hitting the surface per second, which translates to more electrons emitted per second.

Comparing curves $a$ and $b$, it is crystal clear that curve $b$ reaches a higher plateau than curve $a$. This tells us that the saturation current for $b$ is greater than for $a$. Consequently, the intensity of radiation $b$ must be greater than the intensity of radiation $a$:

By carefully dissecting the graph, we've uncovered two crucial pieces of information. From the stopping potential, we deduced that $f_a=f_b$. From the saturation current, we concluded that $I_a<I_b$.

Combining these findings, we arrive at our final answer, which perfectly matches option (a). It's a beautiful example of how a simple graph can elegantly encode the profound quantum nature of light!