Unequal slit intensities reduce interference contrast

Steps:

• Equal initial intensities yield amplitudes a1 = a2 = a, so max intensity (2a)^2 = 4I0, min intensity (0)^2 = 0.
• Reducing one intensity to I1 < I0 gives a1 < a2 = a, so max = (a1 + a2)^2 < 4I0, min = (a2 - a1)^2 > 0.
• Overall pattern: bright fringes dim due to lower max, dark fringes brighten due to nonzero min.
• Contrast (max - min)/(max + min) decreases, confirming reduced visibility.

Why D is correct:

• Interference formula I = I1 + I2 + 2√(I1 I2) cosδ shows unequal I1, I2 yield lower maxima and higher minima than equal case.

Why the others are wrong:

• A: Bright fringes dim, not brighten; darks brighten but not with brights.
• B: Dark fringes brighten, not darken; brights dim but not both.
• C: Bright fringes dim, not brighten; darks brighten, not darken.

Final answer: D