In a biprism experiment, the distance between two coherent sources is 0.5 mm and that between the slit and eyepiece is 1.2m. The slit is successively illuminated by red light of wavelength `6550 Å` and green light of wavelength `5240 Å`. It is found that the nth red bright band coincides with the (n+1)th green bright band. Calcualte the distance of this band from the central bright band.

Explanation

Data: `d= 5 xx 10^(-4)m, D=1.2m, lambda_(r) = 6550 Å= 6.55 xx 10^(-7)`m,
`lambda_(g) = 5240 Å= 5.24 xx 10^(-7)`m,
`x_(n)(red) = x_(n+1)` (green) …………………..[for bright bands]
The distance of the nth bright band from the central bright band is
`x_(n) = nlambdaD/d`
`therefore x_(n) red = nlambda_(r) D/d` and `x_(n+1)`(green)`=(n+1)lambda_(g)D/d`
`therefore nlambda_(r)D/d = (n+1)lambda_(g)D/d`
`therefore nlambda_(r) = nlambda_(g) + lambda_(g)`
`therefore n=lambda_(g)/(lambda_(r)-lambda_(g)) = 5240/(6550-5240) = 5240/1310 =4`
`therefore` The distance of this band from the central bright band is
`x_(4)`(red) `=(4(6.65 xx 10^(-7))(1.2))/(5 xx 10^(-4))`
`=4 xx 1.31 xx 1.2 xx 10^(-3)`
`=6.288 xx 10^(-3) m = 6.288` mm