If the filter is not used over a black light, which problem is likely to occur?

If the filter is not used over a black light, which problem is likely to occur?

• A. No light will be produced
• B. The image will be over-contrasted
• C. Too much white light will be present
• D. The penetrant will fluoresce more brightly

Answer: (C)

The key idea is that a filter on the viewing light is used to block visible light so the fluorescent glow from the penetrant stands out. Fluorescent penetrants emit light when excited by UV, but any white or visible light leaking into the viewing path can wash out that glow and reduce contrast. If the filter isn’t used, white light from the lamp (and ambient light) will mix with the fluorescence, making the defect indications harder to see. So the likely problem is that too much white light will be present, diminishing the visibility of the fluorescence. The other options don’t fit because the lamp still emits UV to excite the penetrant, images aren’t inherently over-contrasted with the filter missing, and the penetrant wouldn’t actually fluoresce more brightly—visibility would just be poorer due to background light.

The key idea is that a filter on the viewing light is used to block visible light so the fluorescent glow from the penetrant stands out. Fluorescent penetrants emit light when excited by UV, but any white or visible light leaking into the viewing path can wash out that glow and reduce contrast. If the filter isn’t used, white light from the lamp (and ambient light) will mix with the fluorescence, making the defect indications harder to see. So the likely problem is that too much white light will be present, diminishing the visibility of the fluorescence. The other options don’t fit because the lamp still emits UV to excite the penetrant, images aren’t inherently over-contrasted with the filter missing, and the penetrant wouldn’t actually fluoresce more brightly—visibility would just be poorer due to background light.