Let $\phi_{0}$ be the work function of the surface of the material. Then,
According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons in the first case is
$$
K_{\max _{1}}=\frac{h c}{\lambda}-\phi_{0}
$$
and that in the second case is
$$
\begin{array}{c}
K_{\max _{2}}=\frac{h c}{\lambda \frac{\lambda}{2}}-\phi_{0}=\frac{2 h c}{\lambda}-\phi_{0} \\
\text { But } K_{\max _{2}}=3 K_{\max _{1}} \text { (given) } \\
\therefore \quad \frac{2 h c}{\lambda}-\phi_{0}=3\left(\frac{7_{1 c}}{\lambda}-0_{0}\right) \\
\frac{2 h c}{\lambda}-\phi_{0}=\frac{3 h c}{\lambda}-30_{0} \\
3 \phi_{0}-\phi_{0}=\frac{3 h c}{\lambda}-\frac{2 h c}{\lambda} \\
2 \phi_{0}=\frac{h c}{\lambda} \text { or } \phi_{0}=\frac{h_{c}}{2 \lambda}
\end{array}
$$
2. (b): Here,
Input signal, $V_{i}=2 \cos \left(15 t+\frac{\pi}{3}\right)$