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Answer :
Here v = +50cm
Because the real image is of the same size as the object,
\(\therefore m=\frac{h`}h=\frac{v}u=-1 \)
or \(u= -v= -50cm \)
Now, \(\frac{1}f=\frac{1}v-\frac{1}u =\frac{+2}{50}=\frac{+1}{25} \)
or \( f=25cm \) or \( 0.25m \)
\( Power(P)=\frac{1}f= +\frac{1}{0.25}=+4D \)