Let's analyze the problem step by step.

• In the first situation, the object is at a distance

$$u_1 = 12 \text{ cm}.$$

Since a virtual image is formed by a convex lens, the image distance is negative (using the sign convention where the object distance is positive and real image distances are positive). Let the image distance be

$$v_1 = -|v_1|.$$

The lens formula is given by

$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}.$$

Thus, for the first case, we have

$$\frac{1}{f} = \frac{1}{v_1} + \frac{1}{12} = -\frac{1}{|v_1|} + \frac{1}{12}.$$

• In the second situation, the object is moved 4 cm further from the lens. Thus,

$$u_2 = 12 + 4 = 16 \text{ cm},$$

and a real image is formed (so its image distance is positive), say

$$v_2.$$

The lens formula now gives

$$\frac{1}{f} = \frac{1}{v_2} + \frac{1}{16}.$$

• We are told that the images have the same size. The magnification is given by

$$m = -\frac{v}{u}.$$

For the magnitudes of the images to be the same, we require that

$$\left|\frac{-v_1}{12}\right| = \left|\frac{-v_2}{16}\right|.$$

Since for the first case we have

$$|v_1| = -v_1,$$

this condition becomes

$$\frac{|v_1|}{12} = \frac{v_2}{16}.$$

Hence,

$$v_2 = \frac{16}{12} |v_1| = \frac{4}{3} |v_1|.$$

• Now, equate the two expressions for $$\frac{1}{f}$$ from the lens formulas:

From the first case:

$$\frac{1}{f} = \frac{1}{12} - \frac{1}{|v_1|},$$

and from the second case:

$$\frac{1}{f} = \frac{1}{v_2} + \frac{1}{16}.$$

Substitute $$v_2 = \frac{4}{3}|v_1|$$ into the second equation:

$$\frac{1}{f} = \frac{1}{\frac{4}{3}|v_1|} + \frac{1}{16} = \frac{3}{4|v_1|} + \frac{1}{16}.$$

Thus, we set:

$$\frac{1}{12} - \frac{1}{|v_1|} = \frac{3}{4|v_1|} + \frac{1}{16}.$$

• Rearranging the equation, subtract $$\frac{1}{16}$$ from both sides:

$$\frac{1}{12} - \frac{1}{16} - \frac{1}{|v_1|} = \frac{3}{4|v_1|}.$$

Compute $$\frac{1}{12} - \frac{1}{16}$$ by finding a common denominator:

$$\frac{1}{12} - \frac{1}{16} = \frac{4}{48} - \frac{3}{48} = \frac{1}{48}.$$

So,

$$\frac{1}{48} - \frac{1}{|v_1|} = \frac{3}{4|v_1|}.$$

• Add $$\frac{1}{|v_1|}$$ to both sides:

$$\frac{1}{48} = \frac{1}{|v_1|} + \frac{3}{4|v_1|} = \frac{1 + \frac{3}{4}}{|v_1|} = \frac{\frac{7}{4}}{|v_1|}.$$

Thus,

$$\frac{1}{48} = \frac{7}{4|v_1|}.$$

Solving for $$|v_1|$$:

$$|v_1| = \frac{7 \times 48}{4} = 7 \times 12 = 84 \text{ cm}.$$

So, the virtual image distance is

$$v_1 = -84 \text{ cm}.$$

• Now, substitute the value of $$|v_1|$$ back into the first lens equation to find the focal length:

$$\frac{1}{f} = \frac{1}{12} - \frac{1}{84}.$$

Find a common denominator:

$$\frac{1}{12} - \frac{1}{84} = \frac{7}{84} - \frac{1}{84} = \frac{6}{84} = \frac{1}{14}.$$

Therefore,

$$f = 14 \text{ cm}.$$

Thus, the focal length of the lens is $$\boxed{14 \text{ cm}}.$$

This corresponds to Option B.

Ray Optics

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