Prove Snell's law of refraction by using Huygens's concept of plane wavefronts.

Explanation

Distance `AE=v_2 tau,tau-` time taken for optical path
Distance `BC=v_1 tau,i_1-` angle of incidence in medium (1)
From the right angled `Delta ABC, sin i=(BC)/(AC)`
And from the right angled `Delta AEC,sin r=(AE)/(AC)`
Hence, `(sini)/(sinr)=(BC)/(AE)=(v_1 tau)/(v_2 tau)`
or `(sin i)/(sin r)=(v_1)/(v_2)" "...(1)`
By definition, absolute R.I of a medium w.r. to air/vacuum `=n= (c )/(v)`
for medium (1) `=n_1-(c )/(v_1)`
for medium `(2)=n_1=(c )/(v_2)`
so that, `(v_1)/(v_2)=(n_2)/(n_1)" "...(2)`
Using (2) in (1) we get
`(sin i)/(sin r)=(n_2)/(n_1)`
where, `n_2 gt n_1`.
or `n_1 sin i=n_2 sin r.................(3)`
i.e. R.I. of medium (1) times since of angle in the medium (1) = R.I of medium (2) times sine of angle in the medium 2.
This expression (3) is known as the Snell's law.
The ratio `(n_1)/(n_1)` can be written as `n_2` (R.I. of medium (2) w.r.t medium (1)).
Note : From (1) , `n_2=(sin i_1)/(sin i_2)=(v_1)/(v_2)=(lambda_1)/(lambda_2)`
i.e. `v_1 sin i_2=v_2 sin i_1,lambda_2 sin i_2 =lambda_1 sin i_2 and sin i_1=n_2 sin i_2`