TS EAMCET 03 May 2019 Shift 2 Solved Paper

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=‌

10×600×10−9

0.3

‌‌=20µm