D

Step 1: We have a convex lens with a focal length (f) of 10 cm. The object distance (u) from one of the pieces of the lens is 15 cm.
Step 2: Using the lens formula, \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \), we can rearrange it to find the image distance (v).
Step 3: For the first piece of the lens, substituting the values: \( \frac{1}{10} = \frac{1}{v_1} - \frac{1}{-15} \) \( \Rightarrow \frac{1}{v_1} = \frac{1}{10} - \frac{1}{15} \).
Step 4: Finding a common denominator (30), we have: \( \frac{1}{v_1} = \frac{3 - 2}{30} = \frac{1}{30} \). So, \( v_1 = 30 \text{ cm} \) (real image on the opposite side).
Step 5: For the second piece of the lens, since it is displaced by 2 cm (1 cm for each piece), the object distance is now \( 15 \text{ cm} - 2 \text{ cm} = 13 \text{ cm} \). Thus, for the second piece:
\( \frac{1}{10} = \frac{1}{v_2} - \frac{1}{-13} \) \( \Rightarrow \frac{1}{v_2} = \frac{1}{10} - \frac{1}{13} \).
Step 6: Again, finding a common denominator (130), we have: \( \frac{1}{v_2} = \frac{13 - 10}{130} = \frac{3}{130} \). Thus, \( v_2 \approx 43.33 \text{ cm} \).
Step 7: The distance between the two images formed is: \( v_2 - v_1 = 43.33 \text{ cm} - 30 \text{ cm} \approx 13.33 \text{ cm} \).
Therefore, the distance between the two images is approximately 13.33 cm. Since we are looking for the options in whole number, the nearest option is 13 cm (if available). Hence, D is correct.