voltage drop across ideal diode
If these are ideal semiconductor diodes (e.g. a pn junction diode), then the current follows the equation I = I0(exp(q.V/kT) -1). As with most electronics there are some approximations here. When reverse biased as D1 is, the reverse current is I0. Typically this is very small (nA).  This flows through D2, and using the equation above results in q.V/kT = ln(2) = 0.69. At room temperature this means that the voltage across D2 is 0.69*kT/q, or about 18 mV. The voltage drop across the 1k resistor will be negligible (uV).