Tungsten filament lamps have a very high positive temperature coefficient of resistance. Typically, the hot resistance is an order of magnitude greater than its cold resistance for ordinary tungsten filaments running in an evacuated envelope such as that festoon lamp.
Halving the current doesn't simply drop the voltage to half, more likely to one quarter or less. Indeed, using that lamp as envisaged, you may not even be able to see it glow at all without peering very closely at the naked lamp whilst shading it from any bright room lighting.
The use of a LED lamp with a fullwave bridge rectifier plus 4 or 5 silicon diodes in series across the DC output and a series current limiting resistor for the LED itself wired across the resulting voltage clipped rectifier output is a far better idea if you don't mind losing 4 or 5 volts from the lamp supply. However, the best version of this "Current Flow Detector" wired in series with the lamp switched live (on ac supplies) is the already proposed use of a current transformer.
If you're going to go to the trouble of using the CT option, choose the turns (or current) ratio carefully so that the indicator LED will glow just as brightly with a 10W LED lamp as it will with that 60W incandescent lamp you mentioned.
Assuming a high quality 10W LED (the 810 Lm equivalent of the American
60W incandescent 120v 750 hour rated lamp, aka the 75W 230v 1000 hour UK/ European equivalent), you'd be looking at a line current of 42mA at unity power factor. That's your worst case condition. In all probability, you'll fit a cheap as chips (fvsvo 'cheap') lamp using a simple "capacitive dropper" ballast circuit with a PF figure the wrong side of the 50% mark resulting in a current more in the region of 80 to 100mA.If we assume the "worst case scenario" from the point of view of guaranteeing a minimum lumens output from our indicator LED then we only need to decide on how much current will prove sufficient to allow the indicator LED to meet this minimum illumination level. Depending on the desired brightness and efficiency of the LED, you're probably looking at an indicator LED current in the region of 1 to 10mA[1].
If you assume the upper figure the CT ratio would need to be a minimum of 4:1 which quite frankly, seems rather a low ratio for a CT. A few years back, I modded the mains sockets fed from my UPS protected supply to provide a "Power Present" indicator lamp. I used vintage (at least two decades old at the time) 3mm dia red LEDS wired in anti-parallel with a cheap 1A silicon diode with a small capacitor and an inrush limiting resistor in series with the LED wired across the Live and neutral.
ISTR using a 47nF cap with a 2K2 half watt resistor for the capacitive volt dropper. When I tentatively checked out the impedance of a 100nF cap by way of a 'sanity check' to jog my memory just now, I was looking at a reactance value at 50Hz of just 32K ohm (just over 7mA's worth if true) so I'm pretty certain I'd have used the more space saving value of 47 or possibly even 33 nF.
Bearing in mind that the LED only sees half of the current cycles in this simple circuit I'm looking at an average current of just 1.75mA (or possibly even just over 1mA with a 33nF cap). These vintage red LEDs provide a more than bright enough indication so I rather suspect a mere one mA will more than suffice using modern higher efficiency indicator LEDs with a fullwave rectifier. If we assume a 1mA current in the secondary of the CT, this allows the use of a 40:1 CT, a more typical turns ratio value for an off the shelf CT imo. :-)
The original circuit (F/W rect with 4 or 5 voltage clamping diodes (or
3v3 zenner diode) feeding a LED via a current limiting resistor) is simply wired across the low current secondary of the CT and the primary wired in series with the live feed to the lamp. Instead of losing 5 volts or so, you'll only lose 125mV across the CT's high current primary. Furthermore, the rectifier and voltage clamping diodes need only handle one fortieth of the lamp current which is a bonus when it comes to coping with short circuit fault currents.You might even be able to make your own CT using a ferrite ring with 20 to 40 turns of fine (28 SWG?) enamelled copper wire wound onto the ring which can then be threaded onto the switched live conductor, neatly avoiding having to insert an extra connector into the circuit. :-)
If you have a suitable ferrite ring and the yard or two of fine enamelled copper wire to hand, you can build a test circuit and test with a suitable lamp holder wired to a 13A plug (I'm assuming the indicator LED and other components won't be a problem for you to source).
There's just one final note: You may find this CT based current indicator possibly a little too sensitive to fully extinguish when the light is switched off. If this turns out to be the case, just wire another resistor (10K?) across the rectifier output pins to divert this 'nuisance current' around the LED part of the circuit.
[1] That was a ball park guess which, on reflection (that memory jogging exercise re the average current in those power present indicator LEDS) was rather on the high side even for vintage LED indicator lamps, let alone the more efficient indicator LEDs available today. If you don't mind the painful brightness of blue LEDs you can reckon on only needing a tenth of the current required by a red LED.