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Use Huygens' principle to show the propagation of a plane wavefront from a denser medium to a rarer medium. Hence find the ratio of the speeds of wavefronts in the two media.

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#### Solution

Let XY be the surface separating the denser medium and the rarer medium.

Let:*v*_{1} = Speed of light wave in the denser medium*v*_{2} = Speed of light wave in the rarer medium

Let us consider a plane wavefront AB propagating in the direction AA'. Let this wavefront incident on the interface at an angle of incidence* i *with the normal to the interface.

Let *τ* be the time taken by the wavefront to travel the distance BC in denser medium.

⇒BC=v1τ

*v*

_{2}

*τ*from point A in the rarer medium. Let CD represent a tangent plane drawn from point C onto the sphere.

Now,

AD=v2τ

Here, CD would represent the refracted wavefront. Considering the triangles ABC and ADC, we get

`sini=(BC)/(AC)=(v_1τ)/(AC)`

` sinr=(AD)/(AC)=(v_2τ)/(AC)`

*i*= Angle of incidence

*r*= Angle of refraction

Since

*r*>

*i*(rays bend away from the normal on travelling from denser to rarer medium), the speed of light in the rarer medium (

*v*

_{2}) will be greater than the speed of light in the denser medium (

*v*

_{1}).

If c represents the speed of light in vacuum, then

`μ1=c/v_1`

` μ2=c/v_2`

where

*μ*

_{1}= Refractive index of denser medium

*μ*

_{2}= Refractive index of rarer medium

Further, (1) can be written as

`μ1sini=μ2sinr`

This is the Snell's law of refraction.

*λ*

_{1}and

*λ*

_{2}represent the wavelengths of light in the denser medium and rarer medium, respectively, and if the distance BC is

*λ*

_{1}, then the distance AD will be

*λ*

_{2}. So, we have

`λ_1/λ_2=(BC)/(AD)=v_1/v_2`

`⇒v_1/v_2=λ_1/λ_2`

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