Question

10.Using Huygens’ geometrical construction of wave front, show how a plane wave g^is reflected from a surface. Hence, verify laws of reflection.​

Answer 1

$\huge\ \pmb{\red{«\: คꈤ \mathfrak Sฬєя \: » }}$

When light is emitted from source the particles around it bounce which are vibrating at same phase is called as wavefront.

Consider any point Q on incident wavefront;

(Attachment)

Suppose wavefront reaches point P' on the refracted wavefront, the disturbance from point Q reaches the point Q' on the refracting surface XY. A'Q'P' represents the refracted wavefront the time taken by light to travel from the point on incident wavefront to the corresponding point or reflected wavefront should always be the same. Now, time taken by light to go from Q TO Q' will be;

$\bold{t = \frac{QK}{c} + \frac{kQ'}{v} -------(i)}$

Where,

• c and we are the velocities of time to mediums

In right angled ∆AQK,<QAK=i

QK=AK din i --------(ii)

In right angled ∆ P'Q'K,<Q'P'K=r

KQ'=KP' sin r -----------(iii)

Subsituting Eq (ii) nd (iii) in Eq (i), we get

$t = \frac{AK \sin(i) }{c} + \frac{KP' \sin(r) }{v} \\ \\ t = \frac{AK \sin(i) }{c} + \frac{(AP' -A K) \sin(r) }{v} \\ \\ t = \frac{(AP')}{v} \sin(r) + ( \frac{ \sin(i) }{c} - \frac{ \sin(r) }{v} )A K \: ---(iv)$

The rays from different points on incident wavefront will take the same time to reach corresponding points on refracted wavefront, ie given by equation 04 is independent of AK. It will happen, so, if

$\frac{ \sin(i) }{c} - \sin( \frac{ r}{v} ) = 0 \\ \\ \frac{ \sin(i) }{ \sin(r) } = \frac{c}{v} \\ \\ \frac{c}{v} = n$

$\: \: \: \: \: \: \: \: \: \: \huge \bold{@WildCat7083 }$