See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

2 1 2 1 n n n n V u R    2 2 1 1 n n n n V R u    u x   2 2 1 1 n n n n v R x    For real image V must be +ve 2 1 1 n n n 0 R x     1 2 1 n R x n n   SECTION – D