The decay of the steady-state electron paramagnetic resonance (EPR) signal from naphthalene tripletstate molecules in ether and other glasses under continuous near-ultraviolet irradiation at 77°K has been examined in some detail. It has been shown that free radicals are produced by sensitized solvent decomposition at a rate proportional to the rate of the disappearance of the triplet-state EPR signal. Also, it has been demonstrated that the latter decay process has a first-order dependence upon naphthalene concentration and that it exhibits a dependence upon excitation intensity, which is consistent with a stepwise two-photon absorption mechanism. The experimental results are discussed in terms of a model that involves the resonant transfer of energy from a highly excited triplet state of naphthalene to the solvent. Excitation of the excited triplet state takes place via the absorption of a photon by a naphthalene molecule in its lowest triplet state. The free radicals are produced by the decomposition of the sensitized solvent molecule. It is believed that the naphthalene molecule is not destroyed in this process but that the EPR signal decay arises from a radical-molecule interaction which effectively shortens the lifetime of the triplet state.