See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

from first refaction 2 1 2 1 v u R      n 1 n 1 v R     n n 1 v R   nR AI v n 1    for 2nd refraction u BI AI 2R      = nR 2R n 1   A C B I I 2R AITS-PT-III-PCM(Sol.)-JEE(Main)/2023 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 8 = (2 n)R (n 1)   and v = R For second surface 2 1 2 1 v u R      1 n 1 n v u R    1 n(n 1) n 1 R (2 n)R R      (2 n)  n(n 1) = (n 1) (n 2) n = 4/3