Wavefront is a locus of all particles of the medium, vibrating in the same phase. 

Huygen's wave theory: 

Consider two media namely, 1 and 2 and, let XY be the surface seperating two media. 
The speed of the waves in these media will be v₁ and v₂. 


Suppose, a plane wavefront AB in first medium is incident obliquely on the boundary surface XY, and its end A touches the second surface at a point A^' at time t = 0 while, the other end B reaches the second surface at point B' after time-interval t.
Here,               BB' = t 

As per Huygen’s principle, secondary spherical wavelets emanates from points A and B , which travel with speed v₁ in the first medium and speed v₂ in the second medium. 

Distance traversed by secondary wavelets in medium 2 in time t , AA' = v₂t 
Distance traversed by point of wavefront in medium 1 in time t = BB' = v₁t 

Now, considering A as the centre we will draw an arc and a tangent( B'A') is drawn from B' to this point. So, as the incident wavefront advances, secondary wavelets start from points in-between A and B' and will reach A'B' simultaneously. 
According to Huygen’s principle A'B' is the new position of wavefront AB in the second medium and, A'B' is the refracted wavefront. 

Let the incident wavefront AB and refracted wavefront A'B' make angles i and r respectively with refracting surface XY. 

In right ,
 
    $\sin i= \sin \angle BAB\text{'} = \frac{BB\text{'}}{AB\text{'}}=\frac{v_{1}t}{AB\text{'}}$    ... (i)

Similarly, in right $∆$AA'B'

   $\sin r = \sin \angle AB\text{'}A\text{'} = \frac{AA\text{'}}{AB\text{'}}=\frac{v_{2}t}{AB\text{'} }$   ...(ii) 

Dividing equations (i) by (ii) , we have  

              $\frac{\sin \mathrm{i}}{\sin \mathrm{r}} = \frac{v_1}{v_2} = cons\tan t$  

which is the required Snell's law.  

Hence proved. 

Given, two convex lenses. 
Aperture of first lens = 5 cm
Aperture of second lens = 10 cm

i) Resolving power is directly proportional to the diameter. 
Therefore, the ratio of resolving power is, 

                    $\frac{\mathrm{R}.{\mathrm{P}}_1}{\mathrm{R}.{\mathrm{P}}_2}= \frac{5}{10}=\frac{1}{2}$

ii) Intensity is directly proportional to the area of objective lens.

$\therefore$                  $\frac{{\mathrm{I}}_1}{{\mathrm{I}}_2} = \frac{5^2}{{10}^2} = \frac{1}{4}$ 

Refractive index of water, ${\mathrm{\mu }}_{\mathrm{air}} = 1$ 
Given, 
Refractive index of glass w.r.t. air, ${}^{\mathrm{a}}\mathrm{\mu }_{\mathrm{g}} = \frac{3}{2}$ 

Refractive inex of water w.r.t air, ${}^{\mathrm{a}}\mathrm{\mu }_{\mathrm{w}} = \frac{4}{3}$  

$\therefore$ refractive index of glass w.r.t. water, = ${}^{\mathrm{w}}\mathrm{\mu }_{\mathrm{g}} = \frac{{}^{\mathrm{a}}\mathrm{\mu }_{\mathrm{g}}}{{}^{\mathrm{a}}\mathrm{\mu }_{\mathrm{w}}} = \raisebox{1ex}{${\displaystyle \frac{3}{2}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \frac{4}{3}}$}\right.$ = $\frac{9}{8}$