Step 1: Use magnification formula for lenses. \[ m = \frac{-v}{u} \] Step 2: Given that image is formed at twice the distance of object from lens.
So, if object distance = \( u \), then image distance \( v = -2u \) (real and on opposite side).
Step 3: Calculate magnification. \[ m = \frac{-(-2u)}{u} = \frac{2u}{u} = 2 \] So, the magnification is \( 2 \). Since the image is real and inverted, but sign conventions were carefully handled, the magnitude remains positive here.
