A vessel ABCD of 10 cm width has two small slits `S_(1)` and `S_(2)` sealed with identical glass plates of equal thickness. The distance between the slits is 0.8 mm. POQ is the line peropendicular 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 form the vessel to illuminate the slits as shown in figure .Calculate the position of the central bright fringe on the other wall CD with respect to line OQ. Now , a liquid is poured into the vessel and filled up to OQ. The central bright fringe is found to be at Q. Calculate the refractive index of the liquid to 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`