In a Young's double slit experiment, the two slits are separated by 0.5 cm and the screen is at 0.5 m from the slits. If 20000 bright fringes are counted per meter on the screen, then the wavelength of light used is

Options

A. `5000overset(@)A`
B. `5890overset(@)A`
C. `6000overset(@)A`
D. `5460overset(@)A`

Answer

A

Explanation

`because " Fringe width, " beta=(Dlambda)/d`,
where, D = distance between screen and slits, d = distance between two slits and `lambda=` wavelength of light, `lambda=(betad)/D`
Given, `d=0.5 times 10^(-2)m, beta=1/20000, D=0.5m`
`therefore lambda=(0.5 times 10^(-2))/(20000 times 0.5)=1/2 times 10^(-6) " or " lambd=5000 overset(@)(A)`