= Solution
A slit of width $w$ has the <single-slit diffraction> <complex amplitude> factor $\sin\alpha/\alpha$, where $\alpha=\pi w\sin\theta/\lambda$. Thus for $w=25\,\mu\mathrm m=d/2$,
$$
\boxed{\frac{I(\theta)}{I(0)}
=\left(\frac{\sin\alpha}{\alpha}\right)^2
\frac{(1+2\cos2\beta)^2}{9},\qquad \alpha=\frac\beta2.}
$$
The <interference> factor keeps the same zeros and the same nominal principal-order positions as in part (i), while the <single-slit diffraction> <wave envelope> suppresses off-axis <wave intensity>. Its zeros are
$$
\sin\theta=p\frac\lambda w=p(0.0200),\quad p\neq0,\qquad
X\simeq p(20.0\,\mathrm{mm}).
$$
At nominal principal <diffraction orders> $\sin\theta_m=m\lambda/d$, the <wave envelope> factor is $[\sin(m\pi/2)/(m\pi/2)]^2$. Consequently
$$
\boxed{\frac{I(\theta_{\pm1})}{I(0)}=\frac4{\pi^2}=0.4053,\qquad
I(\theta_{\pm2})=0,\qquad
\frac{I(\theta_{\pm3})}{I(0)}=\frac4{9\pi^2}=0.0450.}
$$
All nonzero even principal orders are missing. The central <single-slit diffraction> maximum extends between the envelope zeros near $X=\pm20\,\mathrm{mm}$ and includes the central and the $m=\pm1$ principal orders. Because there are only three slits, the maxima of the product can shift slightly from the nominal nonzero-order positions; the quoted values are the exact product at those positions. The lower panel of the figure shows the full product.
Both panels normalize to their own central <wave intensity>. An absolute comparison would require the unspecified widths and transmitted power of the initially “narrow” slits; the normalized pattern change is fully determined.
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