To determine the state of the diodes in the full wave rectifier circuit at $ t = 15 \, \text{ms} $, we analyze the input supply voltage given by:

$ V_{\text{in}} = 220 \sin(100 \pi t) $

Here are the key steps to explain the situation:

Substitute the Time: Convert $ t = 15 \, \text{ms} $ to seconds:

$ t = 0.015 \, \text{s} $

Calculate the Angular Frequency:

$ \omega = 100\pi $

Determine the Period:

$ \frac{2\pi}{T} = 100\pi \quad \Rightarrow \quad T = \frac{1}{50} \, \text{s} = 0.02 \, \text{s} $

Calculate the Phase:

$ t = \frac{3T}{4} $

This corresponds to the negative half-cycle of the sine wave, occurring when the input voltage goes negative.

Given the circuit configuration during the negative half-cycle:

$ D_1 $ is reverse biased.

$ D_2 $ is forward biased.

Hence, at $ t = 15 \, \text{ms} $, the diode $ D_1 $ is reverse biased, and diode $ D_2 $ is forward biased, allowing current to flow through diode $ D_2 $.