What specifically is the effect do you wish to minimize? Perhaps you mean...
1.Tube length changes due to temp change.
2.Condensation on the optics.
To answer the question you asked, the insulation formula for conduction in air is Q=uA(T2-T1), where
Q=rate of heat transfer in BTUH/Hr
A=surface area available for heat transfer in square feet
u=thermal conductivity=1/R
(Units for u are often expressed inconsistently. Much simpler to use "R-value" instead since that's how most insulation is rated anyway. Example: 3.5" thick battes of the pink stuff put between wall studs is R12.)
T2=temp (higher) where heat is flowing FROM in degrees F.
T1=temp (lower) where heat is flowing TO in degrees F.
Keep in mind that insulation only slows heat transfer, it can't stop it altogether. So your 65 degree scope will eventually reach 95 degrees, it's only a matter of time.
Insulation probably won't buy you much time. Wrapping insulation around the barrel not only increases it's R-value, it increases the cylinder's diameter as well. You quickly reach a point where adding more thickness increases the heat transfer surface area faster than it improves resistance to heat flow. You an actually get to a point where you move heat faster than if you had no insulation at all.
Increasing the thermal mass of the device can improve the "time constant" much more effectively (since it has no top end) but it's not practical at all. Essentially, you decide what constitutes a negligle temperature change, decide what time you want that change to occur in, and add enough heat storage capacity to the scope so that the Q above is roughly 100 times less that the total heat stored in the scope's mass. In real world terms, a 1250 pound steel scope with 3 sq.ft. of surface area would take 1 minute to change temp by 1 degree F.
You'd probably be better off just setting the scope outside 15 minutes or so before viewing to let it come to the same temperature as its surroundings.
