= Solution
For a nonabsorbing interface at <normal incidence>, the <Fresnel equations> give the <intensity> <reflectance>
$$
\mathcal R=\left(\frac{n-1}{n+1}\right)^2=0.04.
$$
The two air–glass interfaces each transmit a fraction $1-\mathcal R$ of their incident power. Omitting all internally reflected paths,
$$
\boxed{\frac{I_{\rm transmitted}}{I_{\rm incident}}=(1-0.04)^2=0.9216,\qquad\text{reduction}=7.84\%\simeq8\%.}
$$
The power-transmission factors, rather than electric-field <amplitudes> alone, are multiplied. Absorption, coatings and interference are excluded by this approximation.
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