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Sagot :

Recall the formula of refractive index

[tex]\sf \mu =\dfrac{sini}{sinr}[/tex]

  • i is angle of incidence
  • r is angle of refraction

As per the diagram

At AD angle of incidence i is

[tex]\\ \rm\Rrightarrow \gamma=sin^{-1}\left(\dfrac{1}{\dfrac{n_1}{n_2}}\right)[/tex]

[tex]\\ \rm\Rrightarrow \gamma=sin^{-1}\left(\dfrac{n_2}{n_1}\right)[/tex]

  • That's reverse for BC too

At AB's imaginary boundary line

Use snell's law

[tex]\\ \rm\Rrightarrow \dfrac{\mu_1}{\mu_2}=\dfrac{sin\alpha}{sin\beta}[/tex]

[tex]\\ \rm\Rrightarrow \mu_1sin\alpha=\mu_2sin\beta[/tex]

[tex]\\ \rm\Rrightarrow n_2sin\alpha=n_1sin\left(\dfrac{\pi}{2}-\gamma\right)[/tex]

[tex]\\ \rm\Rrightarrow n_2sin\alpha=n_1cos\left(sin^{-1}\left(\dfrac{n_2}{n_1}\right)\right)[/tex]

  • we need [tex]\sf \alpha[/tex]

[tex]\\ \rm\Rrightarrow sin\alpha=\dfrac{n_1}{n_2}cos\left(sin^{-1}\dfrac{n_2}{n_1}\right)[/tex]

[tex]\\ \rm\Rrightarrow \underline{\boxed{\bf{\alpha=sin^{-1}\left[\dfrac{n_1}{n_2}cos\left(sin^{-1}\dfrac{n_2}{n_1}\right)\right]}}}[/tex]

Option A

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