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
