A vessel ABCD of `10 cm` width has two small slits `S_(1)` and `S_(2)` sealed with idebtical glass plates of equal thickness. The distance between the slits is `0.8 mm`. POQ is the line perpendicular to the plane AB and passing through O, the middle point of `S_(1)` and `S_(2)`. A monochromatic light source is kept at `S, 40 cm` below `P` and `2 m` from the vessel, to illuminate the slits as shown in the figure. Calculate the position of the central bright fringe on the other wall CD with respect of the line `OQ`. Now, a liquid is poured into the vessel and filled up to `OQ`. The central bright fringe is fiund to be at Q. Calculate the refractive index of the liquid.

Answer

2 cm above point Q on side `CD,mu=1.0016`

Explanation

(i) O is the middle point of slits `S_(1)&S_(2)` also `S_(1)S_(2)=d=0.8mm` in figure `tanalpha=(y_(1))/(D_(1))`
`tanalpha=(40)/(200)=(1)/(5)thereforesinalpha=(1)/(sqrt(26))=(1)/(5.1)approx(1)/(5)approxtanalpha`
path difference `Deltax_(1)=SS_(1)-SS_(2)`
`Deltax_(1)=dsinalpha=0.8xx(1)/(5)` or `Deltax_(1)=0.16mm` ...(i)
let R represents the position of CBF. i.e.,
net path difference should be 0
Now `Deltax_(2)=S_(2)R-S_(1)R`
or `Deltax_(2)=dsinthetaimpliesDeltax_(1)=Deltax_(2)` ..(ii)
For central bright fringe
`Deltax_(2)-Deltax_(1)=0impliesdsintheta-Deltax_(1)=0`
`impliesdsintheta=Deltax_(1)=0.16mm`
`implies(0.8)sintheta=0.16impliessintheta=(0.16)/(0.8)=(1)/(5)`
`thereforetantheta=(1)/(sqrt(24))=(1)/(4.9)=(1)/(5)=sintheta`
So, `tantheta=(y_(2))/(D_(2))=(1)/(5)impliesy_(2)=2cm`
(ii).
When liquid of refractive index `mu` is poured
Then for CBF at Q, net path difference =0
`(mu-1)t=Deltax_(1)implies(mu-1)100=0.16impliesmu-1=0.0016`