An acceptable periapical radiograph is produced at 10 mA. If the milliamperage is reduced by half to 5 mA and everything else is unchanged, what must be done to the exposure time to keep the image density the same?

Correct answer: D. Double the exposure time

• A. Leaving the time unchanged halves the mAs and yields a light (underexposed) radiograph.
• B. Quadrupling the time would double the original mAs and give a dark (overexposed) film.
• C. Halving the time as well would cut the mAs to one quarter, producing a badly underexposed, light film.
• D. CORRECT. Density depends on the product mA × time (mAs), and this product obeys the RECIPROCITY LAW: as long as mAs is constant, the density is constant. Halving the mA halves the number of electrons striking the target per second, so the time must be DOUBLED to restore the same total exposure. 10 mA × 0.5 s and 5 mA × 1.0 s both give 5 mAs and identical density.

Note: The tested concept is reciprocity, which is what is presented here. Remember the division of labour: mA and TIME control DENSITY (quantity of photons); kVp controls CONTRAST (quality/energy) and also affects density.