= Two reflected amplitudes and intensity sketch
At the same output in air, use incident intensity $I_{\rm inc}\propto|A|^2$. The two specified output amplitudes, including their common reflection sign, give the <thin-film interference> result
$$
\frac{I_r}{I_{\rm inc}}=|0.15+0.03e^{i\delta}|^2
=\boxed{0.0234+0.0090\cos\delta}.
$$
The output is a cosine with mean $0.0234I_{\rm inc}$, period $2\pi$, and extrema
$$
\boxed{I_{\max}=0.0324I_{\rm inc}\quad(\delta=2\pi m)},\qquad
\boxed{I_{\min}=0.0144I_{\rm inc}\quad(\delta=(2m+1)\pi)}.
$$
The unequal amplitudes prevent a dark minimum; their <interference visibility> is $0.009/0.0234=5/13\simeq0.385$. This is the stipulated two-beam model: the supplied amplitudes already include propagation and transmission factors, and internal re-reflections are excluded by the stated assumption. They need not be recomputed from the later index value.
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