Understand boost converter's voltage conversion ratio
You have this relationship:
$$ \frac{V}{V_g} = 1 + \frac{D1}{D2} $$
For a given \$V_g\$, with D1 held constant, reducing D2 leads to increasing V.

The thing to recognize is that you don't get to make the above choices without changing something else in the circuit. With D1 and \$V_g\$ held constant, the power input remains the same. So the only way you can decrease D2 is to increase the resistance R (i.e. reducing the load) such that the output voltage V rises which gives you a steeper falling inductor output current slope.