MHTCET 17 Apr 2026 Shift 2 Paper

Concept:Maximum fringe contrast is achieved when the intensities of the two coherent sources are equal, making dark fringes completely dark.Explanation:In an interference pattern, the maximum and minimum intensities are written as:$$I_{\max} = I_1 + I_2 + 2\sqrt{I_1 I_2}$$$$I_{\min} = I_1 + I_2 - 2\sqrt{I_1 I_2}$$The contrast between bright and dark fringes is given by the visibility:$$V = \frac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}} = \frac{2\sqrt{I_1 I_2}}{I_1 + I_2}$$For maximum contrast, $V$ must be as large as possible.By the AM–GM inequality, $I_1 + I_2 \geq 2\sqrt{I_1 I_2}$, where equality holds only when $I_1 = I_2$.Therefore, $V$ reaches its maximum value of $1$ only when $I_1 = I_2$.Substituting $I_1 = I_2 = I$ gives:$$I_{\min} = 2I - 2I = 0$$So the dark fringes become perfectly dark, and the bright fringes stay bright, producing the sharpest possible contrast.Answer:The intensities of the two coherent sources should be equal. Hence, the correct option is Option C: equal.