A horizontal ray of light is incident on the right-angled prism with prism angle 6°. If the refractive index of the material of the prism is 1.5, then the angle of emergence will be:

Ray is falling horizontally, i.e., perpendicular
to vertical surface so angle of incidence i₁ =
0; hence, angle of refraction r₁ = 0
We know, for any prism r₁ + r₂ = A.
So, r₂ = 6^o
Again from Snell’s law

$\begin{array}{l}\frac{{\mathrm{sini}}_2}{{\mathrm{sinr}}_2}=\mu \\ \frac{{\mathrm{i}}_2}{{\mathrm{r}}_2}=\mu \end{array}$
as, i₂ and r₂ are small, i₂ = $\mu r_2=1.5\times 6=9^{\circ }$