In Young's double-slit experiment, the slit are 0.5 mm apart and the interference is observed on a screen at a distance of 100 cm from the slits, It is found that the ninth bright fringe is at a distance of 7.5 mm from the second dark fringe from the center of the fringe pattern. The wavelength of the light used is

Options

A. `(2500)/(7)`
B. `2500`
C. `5000`
D. `(5000)/(7)`

Answer

C

Explanation

For bright fringes, `x=(nlamdaD)/(d)`
where `n=0, 1, 2, 3,`……..
For dark fringes, `x=((2n-1)lambdaD)/(2d)`
where `n=1, 2, 3,` ………
As per question, `(9lambdaD)/(d)-(3lambdaD)/(2d)=7.5xx10^(-3) m`
or `(15lambdaD)/(2d)=7.5xx10^(-3) m` or `lamda=(2xx7.5xx10^(-3)xxd)/(15D)`
Substituting the given values, we get
`lambda=(2xx7.5xx10^(-3) m xx0.5xx10^(-3) m)/(15xx1 m)`
`=0.5xx10^(6) m=5000Å`