Given data

• A convex lens forms a real image that is two times magnified.
• Distance between object and image = $45\text{ cm}$.

Magnification relation For a lens, $m = \frac{v}{u}$ Using sign convention for a real image formed by a convex lens:

• object distance $u<0$
• image distance $v>0$

Since the image is real and magnified two times, $m = -2$ Hence, $\frac{v}{u} = -2 \implies v = -2u$

Use object-image separation Object is on the left of lens and real image is on the right, so separation is $|u| + v = 45$ Since $u<0$, $|u|=-u$. Therefore, $-u + v = 45$ Substitute $v=-2u$: $-u + (-2u) = 45$ $-3u = 45$ $u = -15\text{ cm}$ Then, $v = -2(-15)=30\text{ cm}$

Apply lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ Substituting $u=-15$ cm and $v=30$ cm, $\frac{1}{f} = \frac{1}{30} - \left(-\frac{1}{15}\right)$ $\frac{1}{f} = \frac{1}{30} + \frac{2}{30} = \frac{3}{30} = \frac{1}{10}$ Therefore, $f = 10\text{ cm}$

Final answer The focal length of the lens is $\boxed{10}$