The figure shows the interfernece pattern obtained in double slit experiment using light of wavelength 600 nm.
Q. Let `DeltaX_(A)` and `DeltaX_(C)` represent path differences between waves interering at 1 and 3 respectively then `(|DeltaX_(C)|-(|DeltaX_(A)|)` is equal to

Options

A. 0
B. 300 mn
C. 600 nm
D. 900 nm

Answer

B

Explanation

`CBFimplies` path difference `=0`
`DeltaX_(A)toAt" "1`
dark fringe having path difference `=(lamda)/(2)`
`DeltaX_(C)to`At 3,
bright fringes path difference `=lamda`
`therefore(|DeltaX_(C)|-|DeltaX_(A)|)=(lamda-(lamda)/(2))=(lamda)/(2)=(600)/(2)nm=300nm`