The path difference at a point on the screen in young's experiment is `5lamda`. The distance of that point from the central bright band is 0.5 mm. what is the bandwidth?

Options

A. 2.5 mm
B. 1mm
C. 0.1 mm
D. 10 mm

Answer

C

Explanation

The path difference`=5lamda`. It is an integral multiple of `lamda`.
`therefore` The pont is bright, `nlamda=5lamda" "therefore n=5`
the distance of the point from the central bright band
`=nX=5X" "5X=0.5mm" "therefore X=0.1mm`