In passing through a boundary from denser to rarer medium refraction will not take place if-

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light is incident normally on the boundary

the indices of refraction of the two media are same

the boundary is not visible if $μ_{1} = μ_{2}$

angle of incidence is greater than $\sin^{- 1} (μ_{R} / μ_{D})$

answer is C, D.

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Detailed Solution

When light is incident normally on the boundary, then $i = 0$ , So, according to Snell's law $μ_{1} \sin i = μ_{2} \sin r, r$ is also zero, i.e., there is no refraction.

If $μ_{1} = μ_{2}$ , then boundary will not be visible and $r = i$ , i.e., there will be no refraction. If $i > \sin^{- 1} (μ_{R} / μ_{D})$ or $i > i_{c}$ (critical angle), then also there will be no refraction of light and light will be totally internally reflected.