Why is the efficiency of a half wave rectifier equal to 40.6% and not 50%?
The 'efficiency' they are referring to is Conversion Ratio as I found in the wikipedia article about Rectifiers -

Conversion ratio (also called &quot;rectification ratio&quot;, and confusingly, &quot;efficiency&quot;) η is defined as the ratio of DC output power to the input power from the AC supply.

If we go by this convention, assuming transformer and diodes are ideal, and if \$R_L\$ is the load, then &quot;efficiency&quot; would be -
$$e=\frac{P_{dc}}{P_{ac}}=\frac{I_{dc}^2.R_L}{I_{rms}^2.R_L}=\frac{I_{dc}^2}{I_{rms^2}}$$
where \$I_{dc}\$ is the DC component of the current thru  \$R_L\$, and \$I_{rms}\$ is the rms component.
$$\implies e =\frac{(\frac{1}{T}\int_0^TI_m\text{sin}\omega t)^2}{\frac{1}{T}\int_0^TI_m^2\text{sin}^2\omega t}$$
$$\implies e =\frac{\frac{1}{T^2}(\int_0^{T}I_m\text{sin}\omega t)^2}{\frac{1}{T}(\int_0^{T/2}I_m^2\text{sin}^2\omega t+0)}$$
$$\implies e =\frac{\frac{1}{T^2}(\int_0^{T/2}I_m\text{sin}\omega t+0)^2}{I_m^2/4}$$
$$\implies e =\frac{\frac{1}{\omega ^2T^2}([-I_m\text{cos}\omega t]_0^{T/2})^2}{I_m^2/4}=\frac{\frac{1}{\omega ^2T^2}.4I_m^2}{I_m^2/4}$$
putting \$\omega=2\pi/T\$
$$\implies e =\frac{4}{\pi^2}\approx0.405=40.5\%$$