Answer :
[tex] \large \underline{ \underline{ \sf \: Solution : \: \: \: }}[/tex]
Given ,
Focal length = - 2 m
We know that ,
[tex] \large\sf \fbox{ \fbox{Power = \frac{1}{ \: focal \: length \: } }}[/tex]
[tex] \to \sf Power = - \frac{1}{2} \\ \\ \to \sf Power = - 0.5 \: \: D[/tex]
Hence , - 0.5 D is the power of concave lens