The Tyndall effect is observed only when the following conditions are satisfied,
i. The diameter of the dispersed particles is much smaller than the wavelength of the light used.
ii. The diameter of the dispersed particle is not much smaller than the wavelength of the light used.
iii. The refractive indices of the dispersed phase and dispersion medium are almost similar in magnitude.
iv. The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude.

We start by recalling what the Tyndall effect is. It is the scattering of a beam of light by particles present in a colloidal dispersion. Whether this scattering is observed or not depends on two basic physical considerations:

1. The size of the dispersed particles relative to the wavelength of the incident light.

2. The contrast in refractive index between the dispersed phase and the dispersion medium.

Both of these considerations come from classical scattering theory. From Rayleigh’s scattering expression, the intensity of scattered light $$I_s$$ for very small spherical particles is proportional to $$r^6(\Delta n)^2/\lambda^4,$$ where $$r$$ is the radius of the particle, $$\Delta n$$ is the refractive-index difference between particle and medium, and $$\lambda$$ is the wavelength of light in vacuum. We shall use this qualitative formula only to guide the discussion, not to calculate anything.

Condition about particle size:

If the particles are much smaller than the wavelength $$\lambda,$$ the scattering becomes too weak to be seen by the naked eye because the factor $$r^6$$ becomes extremely small. Hence, to observe a visible Tyndall beam, we actually need the particle diameter to be comparable to, or only slightly smaller than, the wavelength. Mathematically, we want

$$r \lesssim \lambda$$ but not $$r \ll \lambda.$$

So statement ii—“The diameter of the dispersed particle is not much smaller than the wavelength of the light used”—is correct. Statement i, which says the diameter is “much smaller,” contradicts the requirement for visible scattering, so statement i is not correct.

Condition about refractive index:

From the same Rayleigh expression the scattering intensity contains the factor $$(\Delta n)^2,$$ where $$\Delta n = n_{\text{particle}} - n_{\text{medium}}.$$ If the refractive indices are almost identical, $$\Delta n \approx 0$$ and the beam is practically unscattered. Therefore, we require a significant difference in refractive indices. This corresponds to statement iv—“The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude.” Statement iii, which says they are “almost similar,” would again suppress scattering and hence is incorrect.

Collecting the correct statements, we have to satisfy:

Statement ii and Statement iv.

Among the given options, Option A lists exactly these two statements.

Hence, the correct answer is Option A.