The refractive index of glass is 1.6 and that of water is 4/3 , both…

The correct answer is Option (4) → (A)-(II), (B)-(IV), (C)-(I), (D)-(III)

The refractive index of glass with respect to water uses the indices of both media with respect to air. If the order of the two media is turned round, the value becomes its reciprocal. For light going from a medium into air, \( \sin i_c=1/n \).

(A) Refractive index of glass with respect to water — This matches (II) \(\frac{6}{5}\). Take air as medium 1, water as medium 2 and glass as medium 3. Then \( n_{32}=n_{31}\times n_{12} \), which gives \( 1.6\times\frac{3}{4} \) or \( \frac{6}{5} \).

(B) Refractive index of water with respect to glass — This matches (IV) \(\frac{5}{6}\). It is the reciprocal of the value in (A).

(C) \(\sin i_c\) for light going from glass into air — This matches (I) \(\frac{5}{8}\), because \( \sin i_c=1/1.6 \).

(D) \(\sin i_c\) for light going from water into air — This matches (III) \(\frac{3}{4}\), because \( \sin i_c=3/4 \) for \( n=4/3 \). Water has the smaller refractive index, so its critical angle is larger than that of glass.

NCERT (2026-27):
Refraction – “If n₂₁ is the refractive index of medium 2 with respect to medium 1 and n₁₂ the refractive index of medium 1 with respect to medium 2, then it should be clear that”

    \( \displaystyle n_{12}=\frac{1}{n_{21}} \)\( \qquad (9.11) \)

“It also follows that if n₃₂ is the refractive index of medium 3 with respect to medium 2 then n₃₂ = n₃₁ × n₁₂, where n₃₁ is the refractive index of medium 3 with respect to medium 1.”

Total Internal Reflection – “The refractive index of denser medium 1 with respect to rarer medium 2 will be n₁₂ = 1/sin i_c. …”