B

Step 1: Find the image distance from the lens using the lens formula: \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)
Here, \( f = 10 \text{ cm} \) (positive for convex lens), \( u = -15 \text{ cm} \) (object distance is negative).
\( \frac{1}{10} = \frac{1}{v} - \frac{1}{-15} \)
Rearranging gives: \( \frac{1}{v} = \frac{1}{10} + \frac{1}{15} = \frac{3}{30} + \frac{2}{30} = \frac{5}{30} = \frac{1}{6} \)
Hence, \( v = 6 \text{ cm} \).

Step 2: Find the position of the image formed by the lens relative to the mirror. The mirror is 10 cm away from the lens, so the distance from the mirror to the image is \( 10 \text{ cm} - 6 ext{ cm} = 4 ext{ cm} \) (this is positive since it is on the same side as the incoming light).

Step 3: Use the mirror formula: \( \frac{1}{f} = \frac{1}{v_m} - \frac{1}{u_m} \)
Here, \( f = -20 \text{ cm} \) (negative for concave mirror) and \( u_m = -4 \text{ cm} \) (object distance for the mirror).
\( \frac{1}{-20} = \frac{1}{v_m} - \frac{1}{-4} \)
Rearranging gives: \( \frac{1}{v_m} = \frac{1}{-20} + \frac{1}{4} = \frac{-1 + 5}{20} = \frac{4}{20} = \frac{1}{5} \)
Hence, \( v_m = 5 \text{ cm} \).

Step 4: The height of the image formed by the mirror can be calculated. The height for the mirror image is given by the magnification: \( m = -\frac{v_m}{u_m} = -\frac{5}{-4} = \frac{5}{4} = 1.25 \).
The height of the image \( h_m = m \times h_o = 1.25 \times 3 ext{ mm} = 3.75 ext{ mm} \).

Final Answer: The length of the image formed by the mirror is approximately 3.75 mm.