1. Let the object distance from the convex lens be $u$ and image distance be $v$.

2. For a lens, magnification is $m = \left|\frac{v}{u}\right|$ Given that the magnitude of magnification is $2$, so $\left|\frac{v}{u}\right| = 2$ This gives two cases:

• Case 1: $v = -2u$ (real inverted image)
• Case 2: $v = 2u$ (virtual erect image)

We now use the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ with $f = 20\text{ cm}$.

3. Case 1: Real image, $v = -2u$

Substitute into lens formula: $\frac{1}{20} = \frac{1}{-2u} - \frac{1}{u}$ $\frac{1}{20} = -\frac{1}{2u} - \frac{1}{u} = -\frac{3}{2u}$ $u = -30\text{ cm}$ So the object distance is $x_1 = 30\text{ cm}$

4. Case 2: Virtual image, $v = 2u$

Substitute into lens formula: $\frac{1}{20} = \frac{1}{2u} - \frac{1}{u}$ $\frac{1}{20} = -\frac{1}{2u}$ $u = -10\text{ cm}$ So the object distance is $x_2 = 10\text{ cm}$

5. Now compute the ratio: $x_1 : x_2 = 30 : 10 = 3 : 1$

Thus, the correct option is A.