Question

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
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 alongside.
Calculate the position of the central bright fringe on the other
wall CD with respect to the line OQ. Now, a liquid is poured
into the vessel and filled upto OQ.The central bright fringe is
found to be at Q. Calculate the refractive index of the liquid.

Answer 1

Given $y_1$=40 cm,$D_1$=2m=200 cm,$D_2$=10 cm
$tan\alpha =\frac{y_1}{D_1}=\frac{40}{200}\frac{1}{5}\Rightarrow \therefore \alpha =ta{n}^{-1}(1/5)$
$sin\alpha =\frac{1}{\sqrt{26}}\approx tan\alpha$
Path difference between S$S_1$ and S$S_2$ is
$\Delta X_1=S{S}_1-S{S}_{2}or\Delta X_1=dsin\alpha =(0.8mm)\left(\frac{1}{5}\right)$
or $\Delta X_1=0.16mm...(i)$
Now, let at point R on the screen, central bright fringe is
observed (i.e., net path difference = 0).
Path difference between $S_2$R and $S_1$R would be
$\Delta X_2=S_{2}R-S_{1}R$
or $\Delta X_2=dsin\theta ...(ii)$
Central bright fringe will be observed when net path
difference is zero.
or $\Delta X_2-\Delta X_1=0$
$\Delta X_2=\Delta X_{1}ordsin\theta =0.16$
or (0.8)sin$\theta =0.16orsin\theta =\frac{0.16}{0.8}=\frac{1}{5}$
$tan\theta =\frac{1}{\sqrt{24}}\approx sin\theta =\frac{1}{5}$
Hence, $tan\theta =\frac{y_2}{D_2}=\frac{1}{5}$
$\therefore y_2=\frac{D_2}{5}=\frac{10}{5}=2cm$
Therefore, central bright fringe is observed at 2 cm above
point Q on side CD.
Alternate solution
$\Delta X$ at R will be zero if $\Delta X_1=\Delta X_2$
or $dsin\alpha =dsin\theta or\alpha =\theta$
or $tan\alpha =tan\theta \Rightarrow \frac{y_1}{D_1}=frac{y}_2{D}_2$
or $y_2=\frac{D_2}{D_1}.y_1=\left(\frac{10}{200}\right)(40)cmor{y}_2=2cm$
The central bright fringe will be observed at point Q. If
the path difference created by the liquid slab of thickness
t = 1 0 cm or 100 mm is equal to $\Delta X_1$, so that the net path
difference at Q becomes zero.
So, $(\mu -1)t=\Delta X_{1}or(\mu -1)(100)=0.16$
or $\mu -1=0.0016or\mu =1.0016$