Slit 1 of a double slit is wider than slit 2, so that the light from slit 1 has an amplitude three times that of the light from slit 2. Show that equation $I=I_{max}{\cos }^2\frac{\varphi }{2}$ is replaced by the equation, $I=\left(\frac{I_{max}}{4}\right)\left(1+3 {\cos }^2\frac{\varphi }{2}\right)$.

$$\begin{gathered} A_1=3A_2 \\ So, I_1=9I_2 \\ Let A_2=A_0 and I_2=I_0 \\ Then, A_1=3A_{0 }and I_1=9I_0 \\ A_{max}=(A_1+A_2)=4A_0 \\ and I_{max}=16I_0 \\ Now, I=I_1+I_2+2\sqrt{I_1{I}_2}\cos \varphi \\ =9I_0+I_0+2\sqrt{9I_0\times I_0}\cos \varphi \\ =10I_0+6I_0 \cos \varphi \\ =10\times \frac{I_{max}}{16}+6\left(\frac{I_{max}}{16}\right)\cos \varphi \\ =\frac{5}{8}{I}_{max}+\frac{3}{8}{I}_{max}\cos \varphi \\ =\frac{5}{8}{I}_{max}+\frac{3}{8}{I}_{max}\left(2 {\cos }^2\frac{\varphi }{2}-1\right) \\ =\frac{1}{4}{I}_{max}+\frac{3}{4}{I}_{max}{\cos }^2\frac{\varphi }{2} \\ =\frac{I_{max}}{4}\left(1+3{\cos }^2\frac{\varphi }{2}\right) \end{gathered}$$