In the figure, assuming the diodes to be ideal:-

The correct answer is Option (1) → \( D_2 \) is forward biased and \( D_1 \) is reverse biased and hence no current flows from B to A and vice-versa

An ideal diode conducts only when its anode is at a higher potential than its cathode (forward bias). The two diodes are in series between A (at \( -15\,\mathrm{V} \)) and B (earthed, \( 0\,\mathrm{V} \)), and their cathodes meet at the common junction.

    \( \text{anode of } D_2 = V_B = 0\,\mathrm{V} \)
    \( \text{anode of } D_1 \text{ (through R)} = -15\,\mathrm{V} \)

\( D_2 \) has its anode at 0 V, the highest potential in the circuit, so it is forward biased and, being ideal, brings the junction to 0 V. Then \( D_1 \) has its cathode at 0 V and its anode at \( -15\,\mathrm{V} \), so it is reverse biased. A series path containing a reverse-biased ideal diode carries no current, in either direction.