CGPET 2017 Solved Paper

It is given d=0.28mm=0.28×10−3m
D=1.4m
The distance of fourth bright fringe from central bright fringe is 1.2cm. In this dance there will be four interference fringes situated. Because the width of bright and dark fringes are same. So,
4β=1.2cm=1.2×10−2m
β= width of interference fringe
β=

1.2×10−2

4

=3×10−3m
β=

λD

d

⇒3×10−3=

λ×1.4

0.28×10−3

⇒λ=

3×10−3×0.28×10−3

1.4

=6×10−7=6000Å