Answer
a) $I=I_{\circ}N^2$
b) $\phi=\pm\frac{2\pi}{N}$
c) pattern gets narrower
Work Step by Step
(a) We know that
$I=I_{\circ}[\frac{sin(\frac{N\phi}{2})}{sin(\frac{\phi}{2})}]^2$
where $\phi=(\frac{2\pi d}{\lambda})sin\theta$ for small $\phi$ $sin(\frac{N\phi}{2})=\frac{N\phi}{2}$ and $sin(\frac{\phi}{2})=\frac{\phi}{2}$
$\implies I=I_{\circ}[\frac{(\frac{N\phi}{2})}{\frac{\phi}{2}}]^2$
$\implies I=I_{\circ}N^2$
(b) We know that the intensity becomes zero when
$sin(\frac{N\phi}{2})=0$
This happens at $\frac{N\phi}{2}=I\pi$
$\implies \phi=\pm \frac{2\pi}{N}$
Thus, $\phi$ ranges between $\phi=\frac{2\pi}{N}$ and $\phi=\frac{-2\pi}{N}$
(c) We know that the value of $\phi$ is inversely proportional to N; when the value of $N$ increases then the value of $\phi$ decreases. The pattern becomes narrower when the number of slits is increased.
