Use a DC coupled oscilloscope.
Then you might burn out the LEDs.
Most half-decent (not cheap Chinese LEDs on e-Bay) LEDs have a detailed datasheet so you can design a circuit that makes then bright but does not burn then out.
use simple Ohm's Law.
A capacitor has nothing to do with an LED unless it is fed a pulsing voltage.
A capacitor has reactance, not resistance. Its reactance limits the current at low frequencies, not at high frequencies. Your capacitor is parallel with the LED so it steals power.
If you want the LED to slowly fade and slowly brighten then the RC must be at the input of the LED driver.
Brightness is not linear, it is logarithmic. 1/10th the power is half as bright. Then you can see in moonlight and in sunlight (disregarding your attenuating iris).
FIRST OFF, I think I need to make the circuit clearer. The 1 ohm resistor is in series with the cathode lead of the LED to monitor the LED current by measuring the IR drop. Since it's a 1 ohm resistor, each millivolt equals 1 milliamp current.
I put the capacitor across the 1 ohm resistor to filter out the spikes and give the voltage measurement not dependent on frequency, assuming the meter was not made for high frequencies. I DID NOT put the capacitor across the LED.
I changed the big cap across the 1 ohm to a 100k from the resistor hot lead to the meter lead, and a .1 uF cap across the meter leads. This RC low pass filter is to do a better job of filtering out the spikes going to the voltmeter.
In my reply, I said that the capacitor has reactance. One thing about reactance is that it is at a particular frequency, But in this JT circuit, the waveform is a pulse which is a series of frequencies, so the capacitor has to be viewed in the time domain, as it charges and discharges with the pulses.
The issue is not apparent brightness to the eye, the issue is the LED output in lumens or candelas, for illumination purposes. My ancient Weston light meter is what I have to measure the LED's light output.
Measuring the voltage with the DC coupled 'scope doesn't tell you what's the amount of power to the LED. It's a complex waveform and you
cannot divide the peak-to-peak voltage of the pulse by 2.8 to get the RMS voltage.
The LED has the same pulses that are across the transistor, collector to emitter. Nothing is between the two, the LED is connected directly to the transistor.
Even with the inaccuracy, perhaps the method I'm using is the best given its ease and low cost.