[TommyJ]

1. Electromagnetic waves consist of discrete, massless units called photons. A photon travels in vacuum at the speed of light.
2. Each photon has energy where is the frequency of the wave and h is a universal constant called Planck’s constant. The value of Planck’s constant is In other words, the electromagnetic waves come in discrete “chunks” of energy.
3. The superposition of a sufficiently large number of photons has the characteristics of a continuous electromagnetic wave.

Photon energy is E=h*f=p*c (where p - is momentum).
Can be derived from E=m*c^2 (photon m=0). In the same time m=m0*y (m0 - rest mass).
In another direction.
y*m0=(E/c^2)*c=E/c
Photon energy E=p*c.=h*f
Hence, photon momentum. p=h*f/c

The "weight"of light is a form of energy and not pounds or kilograms.

[Kryptid]

The "weight"of light is a form of energy and not pounds or kilograms.

You could weigh it that way if you wanted to, though. If you had a container that was perfectly reflective on the inside and filled it with light, it would weigh more than if the container was empty. That's due to E=mc².

[TommyJ]

You could weigh it that way if you wanted to, though. If you had a container that was perfectly reflective on the inside and filled it with light, it would weigh more than if the container was empty. That's due to E=mc2.

What would be the measurement way and value as an instance?
I'd just like to mislead with confusions further.
Thank you.

[Kryptid]

What would be the measurement way and value as an instance?

Putting it on a (sufficiently sensitive) scale would be enough. The weight would depend on how much energy the light has.

[TommyJ]

Putting it on a (sufficiently sensitive) scale would be enough. The weight would depend on how much energy the light has.

It could be possible to direct enough directed flux a sufficiently sensitive scale, right.