Physics - Modern Physics Question with Solution | TestHub

The saturation photocurrent (the maximum constant value of current) depends on the intensity of the incident light. From the given graph, curves $c$ and $d$ reach the same saturation current level. This implies that the intensities of radiations $c$ and $d$ are equal, so $I_c=I_d$.

The stopping potential is the negative potential on the x-axis where the photocurrent becomes zero. The magnitude of the stopping potential is directly proportional to the frequency of the incident light ($e{V}_0=hf-\Phi$). A more negative stopping potential corresponds to a higher frequency.

Observing the x-intercepts of the curves:
Curve $a$ has the least negative stopping potential, meaning it has the lowest frequency $f_a$.
Curves $b$ and $c$ originate from the same point on the negative x-axis, indicating they have the same stopping potential. Therefore, their frequencies are equal, giving $f_b=f_c$.
Since the stopping potential of curve $b$ is more negative than that of curve $a$, its frequency is higher, so $f_b>f_a$.

Combining these observations, we get the relations: $f_b>f_a$, $f_b=f_c$, and $I_c=I_d$.

Answer: ${\mathrm{f}}_{\mathrm{b}}>{\mathrm{f}}_{\mathrm{a}},{\mathrm{f}}_{\mathrm{b}}={\mathrm{f}}_{\mathrm{c}},{\mathrm{I}}_{\mathrm{c}}={\mathrm{I}}_{\mathrm{d}}$