If the focal spot to film distance is increased from 20 cm to 40 cm, how is the intensity of radiation affected?

Get more with Examzify Plus

Remove ads, unlock favorites, save progress, and access premium tools across devices.

FavoritesSave progressAd-free
From $9.99Learn more

Prepare for the ADC Dental Exam. Study ADC Dental Test topics with quizzes and review study materials. Boost your confidence for the test with our comprehensive practice tests and detailed explanations.

The intensity of radiation is influenced by the inverse square law, which states that the intensity of radiation is inversely proportional to the square of the distance from the source. When the distance from the focal spot to the film is increased, the intensity decreases according to this principle.

In this scenario, when the distance is increased from 20 cm to 40 cm, the new distance is twice the original distance. According to the inverse square law, if the distance is doubled, the intensity of the radiation is reduced to one-quarter of its original value. This can be expressed mathematically as:

I2 = I1 / (d2/d1)²

If you substitute the distances, you find:

I2 = I1 / (40 cm / 20 cm)²

I2 = I1 / 2²

I2 = I1 / 4

Thus, the intensity is reduced to one-quarter of the original intensity. This understanding directly aligns with the selected answer regarding the reduction in intensity due to the increased distance.

Subscribe

Get the latest from Examzify

You can unsubscribe at any time. Read our privacy policy