To solve this problem, we need to consider the lens and mirror system.$\\$Given:$\\$• Object distance from the lens $(u) = -60$ cm (since it is on the opposite side of the incident light).$\\$• Focal length of the lens $(f) = 30$ cm.$\\$Using the lens formula:$\\ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \\$Substitute the known values:$\\ \frac{1}{30} = \frac{1}{v} + \frac{1}{60} \\$Rearrange to solve for $v: \\ \frac{1}{v} = \frac{1}{30} - \frac{1}{60} \\ \frac{1}{v} = \frac{2 - 1}{60} \\ \frac{1}{v} = \frac{1}{60} \\ v = 60$ cm$\\$The image is formed 60 cm on the opposite side of the lens.$\\$Now, consider the plane mirror placed 40 cm from the lens.$\\$The distance of the image from the mirror $= 60 - 40 = 20$ cm.$\\$The mirror will form a virtual image 20 cm behind it.$\\$The final image distance from the lens $= 40 + 20 = 60$ cm.$\\$This image is virtual and located 20 cm from the plane mirror.$\\$Therefore, the final image is formed 20 cm from the plane mirror,$\\$and it would be a virtual image. This corresponds to Option 4.