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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 \( 0.25m \)

\( Power(P)=\frac{1}f= +\frac{1}{0.25}=+4D \)

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