因为$$1\textless{}{{\log }_{2}}3\textless{}2$$，所以  $$f\left( {{\log }_{2}}3 \right)=f\left( {{\log }_{2}}3+1 \right)=f\left( {{\log }_{2}}3+2 \right)=f\left( {{\log }_{2}}3+3 \right)={{\left( \frac{1}{2} \right)}^{{{\log }_{2}}3+3}}=\frac{1}{3}\cdot \frac{1}{8}=\frac{1}{24}$$． 
若函数$$f(x)=\begin{cases}{{\left( \frac{1}{2} \right)}^{x}},x\geqslant 4    f(x+1),x\textless{}4 \end{cases}$$，则$$f({{\log }_{2}}3)=$$．

a: $$-\frac{23}{8}$$ 
b: $$\frac{1}{11}$$ 
c: $$\frac{1}{19}$$ 
d: $$\frac{1}{24}$$ 