An ideal transformer is a device that transfers electrical energy from one circuit to another through inductively coupled conductors—the transformer's coils. The primary coil (with $N_P$ turns) is connected to the input voltage ($V_P$), and the secondary coil (with $N_S$ turns) delivers the output voltage ($V_S$).

The relationship between the input (primary) voltage and output (secondary) voltage in a transformer is determined by the ratio of the number of turns in the primary coil to the number of turns in the secondary coil, expressed by the formula:

$\frac{V_S}{V_P} = \frac{N_S}{N_P}$

In this case, the given turns ratio is $$\frac{N_P}{N_S} = \frac{1}{2}$$. To find the correct expression for $\frac{V_S}{V_P}$, we need to invert the given ratio to reflect the relationship with respect to the voltages. This means:

$\frac{N_S}{N_P} = 2$

Therefore, substituting into the voltage ratio equation:

$\frac{V_S}{V_P} = 2$

This indicates that the secondary voltage ($V_S$) is twice the primary voltage ($V_P$). If we were to express the ratio $V_S:V_P$ based on this, it would be:

Option B $2: 1$

This is the correct answer as it represents that, in accordance with the transformer's turns ratio, the voltage in the secondary coil is twice that in the primary coil.