How can it be said that light possesses wave nature?

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(N/A) The wave nature of light is established by the following evidence:
$1$. Maxwell's equations for electromagnetism predicted that light is an electromagnetic wave consisting of oscillating electric and magnetic fields.
$2$. The experimental verification by Heinrich Hertz,who successfully produced and detected electromagnetic waves in the laboratory,confirmed that light behaves as an electromagnetic wave.
$3$. Phenomena such as interference,diffraction,and polarization,which are characteristic of wave motion,can only be explained by considering light as a wave.

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Which one of the following phenomena is not explained by Huygen's construction of wavefront?

$A$ wavefront is the locus of all points where the particles of the medium vibrate with the same:

For light diverging from a point source,

The figure shows a surface $XY$ separating two transparent media,medium-$1$ and medium-$2$. The lines $ab$ and $cd$ represent wavefronts of a light wave travelling in medium-$1$ and incident on $XY$. The lines $ef$ and $gh$ represent wavefronts of the light wave in medium-$2$ after refraction.
$1.$ Light travels as a
$(A)$ parallel beam in each medium
$(B)$ convergent beam in each medium
$(C)$ divergent beam in each medium
$(D)$ divergent beam in one medium and convergent beam in the other medium.
$2.$ The phases of the light wave at $c, d, e$ and $f$ are $\phi_{c}, \phi_{d}, \phi_{e}$ and $\phi_{f}$ respectively. It is given that $\phi_{c} \neq \phi_{f}$.
$(A)$ $\phi_{c}$ cannot be equal to $\phi_{d}$
$(B)$ $\phi_{a}$ can be equal to $\phi_{e}$
$(C)$ $(\phi_{d}-\phi_{c})$ is equal to $(\phi_{f}-\phi_{e})$
$(D)$ $(\phi_{d}-\phi_{c})$ is not equal to $(\phi_{f}-\phi_{e})$
$3.$ Speed of the light is
$(A)$ the same in medium-$1$ and medium-$2$
$(B)$ larger in medium-$1$ than in medium-$2$
$(C)$ larger in medium-$2$ than in medium-$1$
$(D)$ different at $b$ and $d$
Give the answer for questions $1, 2$ and $3$.

$A$ plane wavefront is incident on a water surface at an angle of incidence $60^{\circ}$. It then gets refracted at an angle of $45^{\circ}$. The ratio of the width of the incident wavefront to that of the refracted wavefront is $\left[\sin \frac{\pi}{4}=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}, \sin 60^{\circ}=\frac{\sqrt{3}}{2}, \cos 60^{\circ}=\frac{1}{2}\right]$

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