Which statement is true about high reflectivity surfaces in infrared inspection?

Which statement is true about high reflectivity surfaces in infrared inspection?

• A. They reflect infrared radiation, which lowers emissivity.
• B. They emit infrared strongly.
• C. They absorb infrared completely.
• D. They have no emissivity.

Answer: (A)

In infrared inspection, emissivity is how well a surface emits infrared energy compared to a perfect blackbody. High reflectivity means most of the infrared radiation incident on the surface is reflected rather than emitted. For opaque surfaces, emissivity and reflectivity add up to about one, so when reflectivity is high, emissivity is low. That’s why the statement that these surfaces reflect infrared radiation, which lowers emissivity, is true. The other ideas don’t fit: emitting infrared strongly would require high emissivity, not high reflectivity; absorbing infrared completely would imply no reflection, which isn’t the case for reflective surfaces; and having no emissivity isn’t possible for a real surface at a finite temperature (there’s always some emission).

In infrared inspection, emissivity is how well a surface emits infrared energy compared to a perfect blackbody. High reflectivity means most of the infrared radiation incident on the surface is reflected rather than emitted. For opaque surfaces, emissivity and reflectivity add up to about one, so when reflectivity is high, emissivity is low. That’s why the statement that these surfaces reflect infrared radiation, which lowers emissivity, is true.

The other ideas don’t fit: emitting infrared strongly would require high emissivity, not high reflectivity; absorbing infrared completely would imply no reflection, which isn’t the case for reflective surfaces; and having no emissivity isn’t possible for a real surface at a finite temperature (there’s always some emission).