Question

`\textbf{Can someone help me with this question ? }`


`\textsf{I need proper Explanation}` ​~

Answer 1

Recall the formula of refractive index

`\sf \mu =(sini)/(sinr)`

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

As per the diagram

At AD angle of incidence i is

`\\ \rm\Rrightarrow \gamma=sin^(-1)\left((1)/((n_1)/(n_2))\right)`

`\\ \rm\Rrightarrow \gamma=sin^(-1)\left((n_2)/(n_1)\right)`

• That's reverse for BC too

At AB's imaginary boundary line

Use snell's law

`\\ \rm\Rrightarrow (\mu_1)/(\mu_2)=(sin\alpha)/(sin\beta)`

`\\ \rm\Rrightarrow \mu_1sin\alpha=\mu_2sin\beta`

`\\ \rm\Rrightarrow n_2sin\alpha=n_1sin\left((\pi)/(2)-\gamma\right)`

`\\ \rm\Rrightarrow n_2sin\alpha=n_1cos\left(sin^(-1)\left((n_2)/(n_1)\right)\right)`

• we need
  `\sf \alpha`

`\\ \rm\Rrightarrow sin\alpha=(n_1)/(n_2)cos\left(sin^(-1)(n_2)/(n_1)\right)`

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

Option A