The number of optical isomers possible for a compound is $2^{\mathrm{n}}$ where $\mathrm{n}=$ number of asymmetric carbon atoms. 

As $\mathrm{n}=1$ for $2-\mathrm{chlorobutane}$, it is an optically active molecule with one chiral carbon atom and it forms dextro and laevo isomers.

$2^{\mathrm{n}}=2^1=2$. 

Hence, it has two optical isomers.