A source that is 0.10 m away from a fresnel biprism of angle `2^(@)` which illuminated by the light of wavelength `6280Å`. What will be the width of the fringes(inmm) formed on a screen kept 0.9 m away from the biprism? (Take Refracive index of glass `=1.5pi=3.14`)

Answer

`0.18`

Explanation

The deviation `delta` produced by each half of biprims `=(mu-1)A`
where A is in radian.
Here
`d=2(mu-1)Ax`
`=2(1.5-1)((pi)/180xx2^(@))xx0.1`
`d=(pi)/900m`
`D=0.9+0.1=1m`
Now `beta=(lamdaD)/d`
`beta=((6280xx10^(-10))xx1)/((pi)/900)=1.8xx10^(-4)m=0`
`.18mm`