Answer :

P = 1/f(in m) = 1/-2 = -0.5 D

[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

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