Physics - Geometrical Optics Question with Solution | TestHub

For a prism, the angle of deviation $\delta$ is given by the relation:

$\delta =i+e-A$

where $i$ is the angle of incidence, $e$ is the angle of emergence, and $A$ is the angle of the prism.

For a given deviation, the two possible angles of incidence correspond to $i$ and $e$. Let these be $i_1$ and $i_2$.

Given that the prism is equilateral, $A=60^{\circ }$. The deviation is $\delta =45^{\circ }$.

Substituting these values, we get:

$i_1+i_2=45^{\circ }+60^{\circ }=105^{\circ }$

It is also given that the two angles of incidence differ by $20^{\circ }$:

$i_1-i_2=20^{\circ }$

Adding the two equations:

$2i_1=125^{\circ }\Rightarrow i_1=62.5^{\circ }$

Subtracting the two equations:

$2i_2=85^{\circ }\Rightarrow i_2=42.5^{\circ }$

Thus, one of the angles of incidence is $62.5^{\circ }$.

Answer: $62.5^{\circ }$