Arrange these in ascending order of wavelength while propagating thro…

The correct answer is Option (4) → (C) < (D) < (B) < (A)

When light passes from one medium to another its frequency does not change, but its speed becomes \( v = c/n \). Since \( v = f\lambda \), the wavelength in a medium is

    \( \displaystyle \lambda = \frac{\lambda_0}{n} \)

so the larger the refractive index, the shorter the wavelength. Typical refractive indices:

    \( (A)\ \text{air}: n \approx 1.00 \)
    \( (B)\ \text{water}: n \approx 1.33 \)
    \( (D)\ \text{glass}: n \approx 1.5 \)
    \( (C)\ \text{diamond}: n \approx 2.42 \)

Diamond has the largest refractive index and air the smallest, so in ascending order of wavelength:

    \( (C) \lt (D) \lt (B) \lt (A) \)

NCERT (2026-27):
Refraction of a Plane Wave – “… Further, if λ₁ and λ₂ denote the wavelengths of light in medium 1 and medium 2, respectively and if the distance BC is equal to λ₁ then the distance AE will be equal to λ₂ (because if the crest from B has reached C in time t, then the crest from A should have also reached E in time t); thus,

    \( \displaystyle \frac{BC}{AE} = \frac{\lambda_1}{\lambda_2} = \frac{v_1}{v_2} \qquad (10.7) \)

The above equation implies that when a wave gets refracted into a denser medium (v₁ > v₂) the wavelength and the speed of propagation decrease but the frequency ν (= v/λ) remains the same.”