Question

A ray of light passes from vacuum into a medium of refractive index n, the angle of incidence is found to be twice the angle of refraction. Then the angle of incidence is ?​

Answer 1

$\huge{\bf{\underline{\red{Answer :}}}}$

Let us Consider incident Angle = ∠i

Refrected angel = ∠r

Now given:-

∠i = 2 ∠r

So, ∠r = ∠i / 2

The refrective index of vaccum is and that of the another medium is "n"

Now, according to Snell's law

$\Rightarrow$1 sin ∠ = n sin ∠r

$\Rightarrow$sin ∠i = n sin ( ∠i / 2 )

$\Rightarrow$2 sin ( ∠i / 2 ) cos ( ∠i / 2 )

$\:\:\:\:\:\:\:\:\:\:$= n sin ( ∠i / 2 )

$\Rightarrow$sin ( ∠i / 2 ) [ 2 cos( ∠i / 2 ) - n = 0

$\Rightarrow$sin ( ∠i / 2) = 0 = sin π

$\:\:\:\:\:\:\:\:\:\:$or, ∠i = 2 cos¹(n / 2)

Now, the value of incident Angle Cannot be 2π

So, the value of incident angle is $\boxed{\red{2 cos¹(n / 2)}}$

Answer 2

$\huge\mathrm{Answer}$

Refractive index ( μ ) = 2

Angle of refraction ( r ) = r = sin i / 2

Angle of incidence ( i ) = 2 × angle of refraction ( r )

$\boxed{Snell's \: Law \: ★}$

The ratio of the sine of angle of incidence to sine of angle of refraction is constant for a given pair of medium i.e

$\boxed{μ\: = sin \: i \: \div sin \: r \: }$

⇒ 1 sin ∠ = n sin ∠r

⇒ sin ∠i = n sin ( ∠i / 2 )

⇒ 2 sin ( ∠i / 2 ) cos ( ∠i / 2 ) = n sin ( ∠i / 2 )

⇒ sin ( ∠i / 2 ) [ 2 cos( ∠i / 2 ) - n ] = 0

⇒ sin ( ∠i / 2) = 0 = sin π or, ∠i = 2 cos¹(n / 2)

Now, sin i cannot be sin π

So,

sin i = 2 cos¹( n / 2 )

_________________________________

$\huge\mathbf{★ Notes}$

✍️ 0 / [ 2 cos( ∠i / 2 ) - n ] = 0

✍️ 2 sin ( ∠i ) cos ( ∠i ) = sin 2i

Reason ➡

L.H.S = 2 sin ( ∠i ) cos ( ∠i )

R.H.S = sin 2i

Let sin i = 30°

L.H.S

= 2 sin ( ∠i ) cos ( ∠i )

= 2 × sin 30° × cos 30°

= 2 × 1 / 2 × √3 / 2

[ sin 30° = 1 / 2 , cos 30 ° = √3 / 2 ]

= √3 / 2

R.H.S

= sin 2i

= sin 2 × 30°

= sin 60° [ sin 60° = √3 / 2 ]

= √3 / 2 = L.H.S

Therefore, 2 sin ( ∠i ) cos ( ∠i ) = sin 2i

2 sin ( ∠i / 2 ) cos ( ∠i / 2 ) = sin i

✍️ sin π = 0 [ Does not exist or it's not valid ]