The Young's experiment is performed with the lights of blue `(lambda=4360Å)` and green colour `(lambda=5460Å)`. If the distance of the 4th fringe from the centre is x, then

Options

A. x(Blue)=x(Green)
B. x(Blue)`gt`x(Green)
C. x(Blue)`lt`x(Green)
D. `("x(Blue)")/("x(Green)")=(5460)/(4360)`

Answer

C

Explanation

The distance of the nth bright fringe from the centre is
`x_(n)=(nlamdaD)/(d)" "therefore x_(n)=lamda` (`because` D and d are kept constant)
`therefore (x_("blue"))/(x_("green"))=(lamda_(B))/(lamda_(G))=(4360)/(5460)`
`therefore x_(B) lt x_(G)`
We will get same result, even if we consider the fourth dark fringe.