A ray of light is incident on a 60^{circ} pri | vedclass

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(A) For a prism,the angle of the prism $A$ is given by the relation $A = r_1 + r_2$,where $r_1$ and $r_2$ are the angles of refraction at the first an

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$30$

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(A) For a prism,the angle of the prism $A$ is given by the relation $A = r_1 + r_2$,where $r_1$ and $r_2$ are the angles of refraction at the first and second faces,respectively.In the position of minimum deviation,the ray of light passes symmetrically through the prism,which implies $r_1 = r_2 = r$.Therefore,the relation becomes $A = 2r$.Given that the prism angle $A = 60^{\circ}$,we have $60^{\circ} = 2r$.Solving for $r$,we get $r = \frac{60^{\circ}}{2} = 30^{\circ}$.Thus,the angle of refraction at the first face is $30^{\circ}$.

$A$ ray of light is incident on a $60^{\circ}$ prism at the minimum deviation position. The angle of refraction at the first face (i.e.,incident face) of the prism is.......$^{\circ}$

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