Optics 6 Question 47

49. A vessel ABCD of 10cm width has two small slits S1 and S2 sealed with identical glass plates of equal thickness. The distance between the slits is 0.8mm. POQ is the line perpendicular to the plane AB and passing through O, the middle point of S1 and S2. A monochromatic light source is kept at S,40cm below P and 2m 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.

(2001,5 M)

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Answer:

Correct Answer: 49. 2cm above point Q on side CD,μ=1.0016

Solution:

  1. Given y1=40cm,D1=2m=200cm,D2=10cm

tanα=y1D1=40200=15⇒∴α=tan1(1/5)

sinα=12615=tanα

Path difference between SS1 and SS2 is

ΔX1=SS1SS2 or ΔX1=dsinα=(0.8mm)15 or ΔX1=0.16mm

Now, let at point R on the screen, central bright fringe is observed (i.e., net path difference =0 ).

Path difference between S2R and S1R would be

ΔX2=S2RS1R

 or ΔX2=dsinθ

Central bright fringe will be observed when net path difference is zero.

 or ΔX2ΔX1=0ΔX2=ΔX1 or dsinθ=0.16 or (0.8)sinθ=0.16 or sinθ=0.160.8=15tanθ=124sinθ=15 Hence, tanθ=y2D2=15y2=D25=105=2cm

Therefore, central bright fringe is observed at 2cm above point Q on side CD.

Alternate solution

ΔX at R will be zero if ΔX1=ΔX2

or dsinα=dsinθ

or α=θ

or tanα=tanθ

y1D1=y2D2

or y2=D2D1y1=10200(40)cm

or y2=2cm

(b) The central bright fringe will be observed at point Q. If the path difference created by the liquid slab of thickness t=10cm or 100mm is equal to ΔX1, so that the net path difference at Q becomes zero.

 So, (μ1)t=ΔX1 or (μ1)(100)=0.16 or μ1=0.0016 or μ=1.0016



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