# How efficiency works for objects

Imagine, there is an object(objA) which is not a black body. But this object is a kind of object that does not reflect any energy(light). It can only absorb and transmit.

We know that blackbody emits radiation at the perfect efficiency at different wavelengths.

Now imagine light hits the objA which then transmits this light(hence lets call this reemitting). In this case, why would not this objA have the perfect efficiency at different wavelengths? You might say that objA might absorb some of the light but this absorbed one will cause new re-emit and the transmitted light + re-emitted should still be equal to the efficiency of black body.

Could you explain why not ?

• Can you rephrase? I have a difficult time trying to understand what it is you are asking. Apr 19, 2023 at 8:57
• "which is not a black body. this object is a kind of object that does not reflect any energy(light). It can only absorb and transmit." - isn't that a black body? Apr 19, 2023 at 9:02
• We know that black body cant transmit. Hence it should not be a black body
– Nika
Apr 19, 2023 at 9:05

Closest analog to such construct is black holes, which absorb every type of radiation and (almost) emit none, except for gravitational waves (but for these black-body law does not apply, because it's just space-time curvature and not EM waves). Also there is a Hawking radiation. For this type radiation of black hole,- indeed black-body radiation law applies. For example, black hole with mass $$1~M_☉$$ according to Wien displacement law, $$\lambda _{\text{peak}}={\frac {b}{T}} = \frac {2.90~mK \cdot m}{60~nK} \approx 48~km \tag 1 ,$$
will emit Hawking EM radiation with $$48~km$$ of wavelength. Part of this radiation is absorbed back by a black hole, i.e. goes back to interior of event horizon, but other part - escapes black hole once and for all. So it's not "a perfect non-emitter" as you search for, but at least it's closest construct so far in the universe.