BSc Nursing GNM 2019 Solved Paper

To find the point of coincidence of bright fringes, we can equate the distance of bright fringes from the central maxima, made by both the wavelengths of light. Given, distance between the screen and slit,
D=120cm=120×10−2m
Slit width,d=2mm=2×10−3m λ1=6500Å and λ2=5200Å
Let nth fringe due to λ2=5200Å coincide with (n−1) th bright fringe due to λ1=6500Å
∴‌‌‌

Dnλ2

d

=‌

D(n−1)λ1

d

‌ or ‌n×5200=(n−1)6500
‌ or ‌4n=5n−5
∴‌‌n=5
So, the least distance required,
yn‌‌=‌

nλ2D

d

‌‌=‌

5×5200×10−10×120×10−2

2×10−3

‌‌=0.16×10−2m
‌‌=0.16cm