The angle of minimum deviation (\(D\)) for a prism is determined by the prism formula: \( n = \frac{\sin((A + D)/2)}{\sin(A/2)} \), where \(n\) is the refractive index and \(A\) is the prism angle. For an equilateral prism with \(A = 60^\circ\) and a refractive index \(n = \sqrt{2}\), we aim to find \(D\). Substituting the given values into the formula yields: \( \sqrt{2} = \frac{\sin((60^\circ + D)/2)}{\sin(60^\circ/2)} \). Since \( \sin(30^\circ) = \frac{1}{2} \), the equation becomes: \( \sqrt{2} = 2\sin((60^\circ + D)/2) \). Rearranging gives: \( \sin((60^\circ + D)/2) = \frac{\sqrt{2}}{2} \). Recognizing that \( \frac{\sqrt{2}}{2} = \sin(45^\circ) \), we have: \( (60^\circ + D)/2 = 45^\circ \). Solving for \(D\), we get: \( 60^\circ + D = 90^\circ \), which simplifies to \( D = 90^\circ - 60^\circ = 30^\circ \). Therefore, the angle of minimum deviation is 30°.