Rather complex problem of odds

Here is a situation ––

you hold two spades, and two spades appear on the flop.
Your odds of drawing a flush on either turn or river are thus roughly 2:1.

The turn comes on a diamond.

The chances of you drawing your spade on the river are now....

4:1, or do they remain 2:1 ?

Simple logic dictates that the chances of drawing the spade on only one card are 4:1, yet since this is the second of two chances, the odds are surely higher in your favor, and seem to me to remain 2:1?

If I'm wrong and the odds are actually 4:1, demonstrate why, but I've been going through a few books (most recently Flynn's Professional No-Limit), and this problem is ignored, and the odds in this situation were understood as 4:1, thus ignoring the context.

Flush draw on the flop:

9 outs
47 cards left
38/47 chance of not getting the flush on the turn
37/46 chance of not getting the flush on the river after not getting it on the turn.

38/47*37/46 = .6503237743

1-probability of not hitting the flush by the river= probability of hitting the flush by the river
1-.6503237743=.3496762257


Good luck.

Edit:
The math above shows how to get 34% which is where 2:1 came from. It's the odds of picking up a flush by the river(hitting a same suit card on either the turn or the river or both)

When the turn comes and it isn't a completed flush, your odds drop cuz there's one less card. It's now 9/46. Approximately 19.4% aka 4:1.

The odds of 2:1 only apply if you plan to see both turn and river..If you only want to see the turn then it will be 4:1.