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)gtX_(green)`
C. `X_(blue)ltX_(green)`
D. `X_(blue)//X_(green)`

Answer

C

Explanation

In young double slit experment position of bright fringes on the screen are given vby
`x=(nDlambda)/(d)`
Hence distance of ntjh maxima will be
`X=(nlambda(D))/(d) proplamdba rArr lambda_(blue)=4360 A rArr lambda_(green)=5480A`
X=distance of 4th maxima from the central are
As `lamdba_(blue)ltlambda_(green)rArrx_(blue)ltx_(green)`