The following figure shows the diagram for Young’s double slit experiment
Therefore, the path difference introduced in slit 1 is given by, ∆x1‌‌=(µ1−1)t ‌‌=(1.6−1)t ‌‌=0.6t The path difference introduced in slit 2 is given by, ∆x2‌‌=(µ2−1)t ‌‌=(1.3−1)t ‌‌=0.3t Therefore, the net path difference introduced in central maxima is given by, ∆x‌central maxima ‌‌‌=∆x1−∆x2 ‌‌=0.6t−0.3t ‌‌=0.3t Calculate the value of t for central maxima, which occupied the 10‌th ‌ bright fringe. ∆xcen‌‌=10λ t‌‌=‌