Dear Mehdi,

in your circuit there are three points, which are disadvantageous: Inputs are too low impedant and transistor base emitter junctions are zenering when being reverse biased, means if signal at 'tip' and 'ring' terminal shows wrong polarity. Also, there must be a switch to disconnect the battery from circuit, otherwise always turned-on LED 2 is discharging the battery.

I would recommend you this circuit below:



It uses two identical sections for each polarity of input signal, is very high impedant at inputs, is protected against surges and bursts and provides a battery test function. Only disadvantage is that it needs two 9V batteries.

Surge protection is provided by varistor S10K150 (Epcos).
BZX55-C6V8 is a small signal zener diode, which limits the level at gate of BS170 to a sane value.
Diode 1N4004 provides a protection of BS170, if signal of wrong polarity is applied. Also, this diode increases the potential needed to turn-on the BS170. So, about 0.6V + 2.0V = 2.6V is needed to turn-on the according LED.
100nF X7R capacitor provides a smooth rise at gate of BS170 if a voltage spike is at 'tip' and 'ring' terminals. Also, it provides a suppresion of hum, as the 1M resistor in combination with 100nF capacitor form a low pas filter with a corner frequency of about 2Hz. So, 50Hz hum is damped by about 30dB!
The 4M7 resistor helps to discharge the 100nF, if once charged. This in order to make the LED quickly turn-off if signal is no longer being applied.
Section on left hand side additionally provides a biasing scheme for the left BS170, which is only avtive when the pushbutton is pressed. This biasing is needed to check for the conditions 'cable break' and 'cable short-circuit'. The biasing also provides the ability of a battery check.

The whole circuit works like that:

Before connecting the tester to phone cable, press the pushbutton. This will bias the left BS170 and make the red LED to turn-on. This will provide you a simple battery check.
Afterwards, connect the phone cable to the 'tip' and 'ring' terminals of tester. If the correct polarity was applied, then the right BS170 turns-on and the green LED begins to shine.
If the wrong polarity was applied, then the left BS170 turns-on and the red LED begins to shine.
If neither the green nor the red LED will shine, then there's either a 'cable break' or a 'cable short-circuit'. For testing this, press the pushbutton. If there's a 'cable short-circuit', then the biasing of the left BS170 is also short-circuited and the red LED stays turned-off. But if there's a 'cable break', then the left BS170 becomes biased and the red LED turns-on.

An additional note: If this circuit is too high impedant, means if you suffer too much from hum effects, then you can decrease the high ohmic resistor values a bit. But keep the ratio unchanged!

Kai