Calculate the angle of minimum deviation for a ray which is refracted through an equiangular prism of refractive index 1.4

Answer Details

The formula for the angle of minimum deviation for an equiangular prism is given as: $$ \delta_{\text{min}} = \frac{A}{2} + \text{arcsin}\left(\frac{n}{\sqrt{2}}\sin\frac{A}{2}\right) - 90^\circ $$ where $A$ is the angle of the prism and $n$ is the refractive index of the prism. For an equiangular prism, $A = 60^\circ$. Substituting $A$ and $n=1.4$ into the formula, we get: $$ \delta_{\text{min}} = \frac{60^\circ}{2} + \text{arcsin}\left(\frac{1.4}{\sqrt{2}}\sin\frac{60^\circ}{2}\right) - 90^\circ \approx 29^\circ $$ Therefore, the angle of minimum deviation for a ray refracted through an equiangular prism of refractive index 1.4 is approximately 29 degrees. Thus, the correct option is 290.