To solve this problem, we will use the mirror formula:$\\ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \\$where $f$ is the focal length, $v$ is the image distance, and $u$ is the object distance.$\\$Initially, the object is placed at $u_1 = -40$ cm (negative sign because the object is in front of the mirror).$\\$The focal length of the concave mirror is $f = -15$ cm.$\\$Using the mirror formula:$\\ \frac{1}{-15} = \frac{1}{v_1} + \frac{1}{-40} \\$Rearrange to solve for $v_1: \\ \frac{1}{v_1} = \frac{1}{-15} + \frac{1}{40} \\$Calculate the right-hand side:$\\ \frac{1}{v_1} = \frac{-8 + 3}{120} = \frac{-5}{120} = \frac{-1}{24} \\ v_1 = -24$ cm$\\$Now, the object is displaced 20 cm towards the mirror, so the new object distance is:$\\ u_2 = -40 + 20 = -20$ cm$\\$Using the mirror formula again:$\\ \frac{1}{-15} = \frac{1}{v_2} + \frac{1}{-20} \\$Rearrange to solve for $v_2: \\ \frac{1}{v_2} = \frac{1}{-15} + \frac{1}{20} \\$Calculate the right-hand side:$\\ \frac{1}{v_2} = \frac{-4 + 3}{60} = \frac{-1}{60} \\ v_2 = -60$ cm$\\$The displacement of the image is the difference between the final and initial image distances:$\\ \Delta v = v_2 - v_1 = -60 - (-24) = -60 + 24 = -36$ cm$\\$The negative sign indicates that the image moves away from the mirror.$\\$Therefore, the displacement of the image is $36$ cm away from the mirror, which corresponds to Option 2.