To add to Stevens answer and give a basic circuit, here is one using LEDs with a Vf of ~2V, based around an N-channel MOSFET:
This is based on a typical LDR resistance range of 12kΩ (light) to 200kΩ (dark). If your LDR covers a different range you will have to alter the resistance values. You may want to add a cap (e.g. >1uF) between gate and ground to filter any noise present and ensure LEDs don't flicker.
R4 and R5 form a voltage divider to bring the 48V down to a bit less than 9.6V (as R4 is in parallel with R2 + R3) at the top of R2. Most MOSFETs don't like too high a voltage on their gate (MOSFET shown actually has 20V Vgs max), so keeping it under 10V is a good idea. Also make sure the MOSFET you use is rated for above 48V drain to source breakdown voltage (this one is rated at 75V). One more thing, given the highish voltage, you must be careful to ensure none of the components dissipate more power than they should.
Then R2 and the LDR form a divider to control the gate voltage. The gate threshold (turn on) voltage of this the IRFP2907 is between 2V(min) and 4V(max), so we want the gate voltage to pass through 4V about halfway through the LDR range.
So a rough calculation, assuming ~9V at the top of R2, and 100kΩ for LDR halfway:
(100kΩ / (4.5V / 9V)) - 100kΩ = 100kΩ needed for R2
To check -> 9V * (100k / (100k + 100k) = 4.5V
Using these values, here is a simulation of the LED current over the LDR resistance range:
The LEDs switch on when the LDR reaches around 90kΩ, which is close enough. You could use a pot for R2 if you want to have the ability to adjust at what light level the LEDs turn on.