How this calculation works
A transparent interface can reflect part of an incident beam even when neither material absorbs light. Fresnel coefficients connect the electric-field amplitudes and the optical power on each side. This page evaluates normal incidence. Power and field amplitude are deliberately reported separately. Squaring the transmitted electric-field amplitude alone is not enough because the power carried by that field also depends on refractive index.
R = [(n1−n2)/(n1+n2)]²; T = 4n1n2/(n1+n2)²; R+T = 1A worked example
For indices one and 1.5, the electric reflection amplitude is −0.2 and power reflectance is 0.04. Electric transmission amplitude is 0.8, while power transmittance is 0.96. An incident 10 W beam therefore gives 0.4 W reflected and 9.6 W transmitted. If the two indices are equal, reflectance becomes zero and all entered power transmits in this ideal model.
Good to know
Only one interface is included. A finite plate has another interface and can introduce multiple reflections or coherent interference, so its total throughput is not automatically this single-surface answer. Absorbing materials, coatings, oblique angles and roughness are absent. The calculation does not estimate scatter, heating or detector response; it is a lossless boundary power balance.
Reference: OpenStax — electromagnetic wave interfaces (opens in a new tab) ↗
What this calculation answers
A transparent interface can reflect part of an incident beam even when neither material absorbs light. Fresnel coefficients connect the electric-field amplitudes and the optical power on each side. This page evaluates normal incidence.
Power and field amplitude are deliberately reported separately. Squaring the transmitted electric-field amplitude alone is not enough because the power carried by that field also depends on refractive index.
The Snell refraction critical angle and Brewster angle calculator changes ray direction. At normal incidence the direction stays on axis while reflected and transmitted powers can still differ.
Use the right input conventions
Enter real refractive indices for transparent nonmagnetic media at a common wavelength. Incident optical power is a nonnegative value in watts. Zero incident power is allowed and gives zero reflected and transmitted watts.
Normal incidence means the beam travels along the interface normal. At this angle the two linear polarization cases coincide, so no separate s or p selection is needed.
Interpret the result
The reflected power fraction is the square of the field reflection coefficient. The transmitted fraction contains an additional refractive-index ratio, giving R + T equal to one for the lossless interface.
A negative electric reflection amplitude denotes a phase reversal in the chosen convention, not negative reflected power. A field transmission amplitude above one likewise does not imply creation of energy.
Work through the example
For indices one and 1.5, the electric reflection amplitude is −0.2 and power reflectance is 0.04. Electric transmission amplitude is 0.8, while power transmittance is 0.96.
An incident 10 W beam therefore gives 0.4 W reflected and 9.6 W transmitted. If the two indices are equal, reflectance becomes zero and all entered power transmits in this ideal model.
Where the model stops
Only one interface is included. A finite plate has another interface and can introduce multiple reflections or coherent interference, so its total throughput is not automatically this single-surface answer.
Absorbing materials, coatings, oblique angles and roughness are absent. The calculation does not estimate scatter, heating or detector response; it is a lossless boundary power balance.
A few questions, answered
Why is transmittance not simply 0.8 squared in the example?
The transmitted wave carries power in a different-index medium, requiring the index factor as well as squared field amplitude.
What changes in the comparison?
Match both indices and inspect the zero-reflection limit.