Formula for efficiency of half wave rectifier
Half wave is 50% max with no load
So 40% means 80% of peak or a drop of 20% from impedance ratio with ripple and DC drop. Optimum depends on ripple tolerance and transformer losses typically design for 10% for some acceptable ripple .
Since charging duty cycle is limited to only when AC-pk&gt;DC out, the peak current/avg DC current is about the same ratio in order to maintain &lt;=10% ripple, this brings down efficiency another bit with cap ESR and diode Rs.
But how they defined these assumptions determines the optimum ratio.