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Prove Snell's law of refraction by using Huygens's concept of plane wavefronts. |
Answer» SOLUTION : 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` |
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