In young's double slit experiment using a source of light of wavelength 5000 `overset(@)A`, the bandwidth obtained is 0.6 cm. If the distance between the screen and the slit is reduced to half, what should be the wavelength of the source to get fringes 0.003m wide?

Explanation

Given `lambda=5000 overset(@)A = 5xx10^(-7)m`
`rArr beta_1=0.06 cm=6xx10^(-2)m`
`rArr beta_2=0.003 m = 3xx10^(-3) m, D_2=(D_1)/(2)`
So that, `(beta_2)/(beta_1)=(lambda_2 D_2)/(lambda_1 D_1)=(lambda_2)/(lambda_2)xx(1)/(2)`
`therefore lambda_2+(2beta_2 lambda_1)/(beta_2)=(2xx3xx10^(-3)xx5xx10^(-7))/(6xx10)m`
i.e. `lambda_2=5xx10^(-3) m = 5000 overset(@)A`
wavelength of light should be 5000 `overset(@)A`.