why if we double the current through an LED, its optical output power approximately doubles?

Question

if we double the current through an LED, its optical output power approximately double

Answer 1

The voltage drop across the diode is fixed, usually around .6V. The power used by a circuit element is:

P = IV

The power used by the diode goes to the production of light. Since V stays the same:

P1 = IV

P2 = 2IV

So, P2/P1 = 2. The diode uses twice the power when the current is doubled.

Answer 2

What we have listed are the official quadruple doubles recorded. But for sure, Wilt Chamberlain and Bill Russel must also have gotten more than a few quadruple doubles in their day as they were awesome rebounders and shot blockers and Wilt even led the league in assists once, while Bill was a good passing center too. Obviously, they could score, especially Wilt. But they didn't keep blocked shots records then. But old time greats like Jerry West, Oscar Robertson, John Havlicek and Elgin Baylor keep insisting that they would so often get like 10-12 blocks in a game! They were that dominant defensively. Maybe Jerry West also got a quadruple double (including steals, aside from pts, rebs and assists) sometime in the 60's, but they didn't yet officially record steals either.

Answer 3

Because if the current doubles, the number of promoted electrons doubles, and twice as many photons are created as the promoted electrons return to the ground state.