Consider two isosceles prisms 1 and 2 with prism angles A_1 and A_2 and refractive indices n_1 and n_2, respectively, as shown in the figure. The faces a_1 b_1 and a_2 b_2 are parallel to each other and perpendicular to the mirror M. If a ray of light is incident on the face a_1 c_1 and emerges from the face a_2 c_2, then the correct statement(s) is/are:
D. If prism 1 is at minimum deviation condition, then i_2 = n_1 ( A_1 2 ) is always true.
From the given geometry, faces a_1b_1 and a_2b_2 are parallel to each other and perpendicular to the horizontal mirror M. Thus, a_1b_1 and a_2b_2 are vertical. The normal to a_1b_1 is horizontal. The ray emerging from Prism 1 makes an angle e_1 with this horizontal normal. By the law of reflection at mirror M, the reflected ray also makes an angle e_1 with the horizontal. This reflected ray is incident on the vertical face a_2b_2 of Prism 2. The normal to a_2b_2 is horizontal, so the angle of incidence is i_2 = e_1. Evaluating option (4): If Prism 1 is at minimum deviation, i_1 = e_1. The condition for minimum deviation gives i_1 = n_1 ( A_1 2 ). Since i_2 = e_1, it follows that i_2 = i_1, yielding i_2 = n_1 ( A_1 2 ). Statement (4) is correct. Evaluating option (1): If both prisms are at minimum deviation, i_1 = e_1 and i_2 = e_2. Thus i_1 = e_1 = i_2 = e_2. Using the minimum deviation formula, i_1 = n_1 ( A_1 2 ) and i_2 = n_2 ( A_2 2 ). Equating them gives n_1 ( A_1 2 ) = n_2 ( A_2 2 ) n_2 n_1 = ( A_1 2 ) ( A_2 2 ) . Statement (1) is correct. Evaluating option (2): If only Prism 2 is at minimum deviation, i_2 = n_2 ( A_2 2 ). Since i_2 = e_1, we get e_1 = n_2 ( A_2 2 ), which does not imply i_1 = n_2 ( A_2 2 ) in general. Statement (2) is incorrect. Evaluating option (3): The dashed lines in the figure bisect the prism angles A_1 and A_2. Since faces a_1b_1 and a_2b_2 are vertical, the angle bisector of Prism 1 makes an angle A_1 2 with the vertical, and the angle bisector of Prism 2 makes an angle A_2 2 with the vertical. The angle between these bisectors is = A_1 2 + A_2 2 . For thin prisms at minimum deviation, the deviation is _m = (n - 1)A, giving A_1 = _ m1 n_1 - 1 and A_2 = _ m2 n_2 - 1 . Substituting these into the expression for , we get = _ m1 2(n_1 - 1) + _ m2 2(n_2 - 1) . Statement (3) is correct.
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