Challenges in optics

Using formula $\color{#3D99F6}{\delta} = \color{#20A900}{i_1} + \color{#D61F06}{i_2} - \color{#69047E}{ A}$
where $\color{#3D99F6}{\delta}$ is angle of deviation, $\color{#20A900}{i_1}$ is angle of incidence on face 1, $\color{#D61F06}{i_2}$ is angle of incidence on 2 and $\color{#69047E}{A}$ is angle of prism.

$30 = 60 + \color{#D61F06}{i_2} - 30 \Rightarrow \color{#D61F06}{i_2} = 0$ Hence, angle of refraction on second face is zero ( $\color{#0C6AC7}{r_2} = 0$ ).

By the formula, $\color{#69047E}{A} = \color{#624F41}{r_1} + \color{#0C6AC7}{r_2}$
where $\color{#624F41}{r_1}$ is angle of refraction on first face of prism.
$\color{#624F41}{r_1} = 30$ Applying Snell's law on face 1,
$\sin \color{#3D99F6}{i_1} = \color{#EC7300}{\mu }\sin\color{#624F41}{ r_1}$ $\sin 60 =\color{#EC7300}{ \mu} \sin 30 \Rightarrow \color{#EC7300}{\mu} = \sqrt{3}$