A convex lens of focal length 20 cm produces images of the same magnification 2 when an object is kept at two distances $$x_1$$ and $$x_2$$ ($$x_1 > x_2$$) from the lens. The ratio of $$x_1$$ and $$x_2$$ is:

We have a thin convex lens whose focal length is given as $$f = +20\ \text{cm}$$ (positive sign because a convex lens is converging).

For any lens the lens (Gaussian) formula is first stated:

$$\frac1f = \frac1v - \frac1u$$

where $$u$$ is the object distance (measured from the lens, taken negative for a real object on the incident-light side) and $$v$$ is the image distance (taken positive for a real image, negative for a virtual image).

The linear (lateral) magnification produced by the lens is defined as

$$m = \frac{h_i}{h_o} = \frac v u.$$

According to the question the magnitude of magnification is the same in two different situations and its value is $$|m| = 2$$. Thus, one situation corresponds to $$m = -2$$ (real, inverted image) and the other to $$m = +2$$ (virtual, erect image). Let the corresponding object distances be $$u_1$$ and $$u_2$$ and their magnitudes be $$x_1$$ and $$x_2$$ respectively. The condition given is $$x_1 > x_2$$, so naturally $$|u_1| > |u_2|$$.

First we treat the case $$m_1 = -2$$.

Because $$m_1 = \dfrac{v_1}{u_1} = -2,$$ we can write

$$v_1 = -2u_1.$$

Substituting this relation into the lens formula, we obtain

$$\frac1f \;=\; \frac1{v_1} - \frac1{u_1} \;=\; \frac1{-2u_1} - \frac1{u_1}.$$

Simplifying the right-hand side step by step,

$$\frac1{-2u_1} - \frac1{u_1} \;=\; -\frac1{2u_1} - \frac1{u_1} \;=\; -\frac1{2u_1} - \frac{2}{2u_1} \;=\; -\frac{3}{2u_1}.$$

Equating this to $$\dfrac1f$$ and putting $$f = +20\ \text{cm}$$ gives

$$\frac1{20} = -\frac{3}{2u_1}.$$

Cross-multiplying,

$$2u_1 = -3 \times 20,$$

$$u_1 = -\frac{60}{2} = -30\ \text{cm}.$$

The magnitude of the object distance is therefore

$$x_1 = |u_1| = 30\ \text{cm}.$$

Now we analyse the case $$m_2 = +2$$.

Because $$m_2 = \dfrac{v_2}{u_2} = +2,$$ we have

$$v_2 = 2u_2.$$

Substituting in the lens formula once again,

$$\frac1f = \frac1{v_2} - \frac1{u_2} = \frac1{2u_2} - \frac1{u_2}.$$

Simplifying term by term,

$$\frac1{2u_2} - \frac1{u_2} = \frac1{2u_2} - \frac{2}{2u_2} = -\frac1{2u_2}.$$

Setting this equal to $$\dfrac1{20}$$ yields

$$\frac1{20} = -\frac1{2u_2}.$$

Cross-multiplying,

$$2u_2 = -20,$$

$$u_2 = -10\ \text{cm}.$$

Thus the magnitude of the second object distance is

$$x_2 = |u_2| = 10\ \text{cm}.$$

Finally we determine the required ratio:

$$\frac{x_1}{x_2} = \frac{30}{10} = 3 : 1.$$

Hence, the correct answer is Option D (3 : 1).