Optics 6 Question 53

55. A double slit apparatus is immersed in a liquid of refractive index 1.33. It has slit separation of 1mm and distance between the plane of slits and screen is 1.33m. The slits are illuminated by a parallel beam of light whose wavelength in air is 6300\AA.

(1996,3M)

(a) Calculate the fringe width.

(b) One of the slits of the apparatus is covered by a thin glass sheet of refractive index 1.53. Find the smallest thickness of the sheet to bring the adjacent minimum as the axis.

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Answer:

Correct Answer: 55. (a) 0.63mm (b) 1.579μm

Solution:

  1. Given, μ=1.33,d=1mm,D=1.33m,

λ=6300\AA

(a) Wavelength of light in the given liquid :

λ=λμ=63001.33\AA4737\AA=4737×1010m

Fringe width, ω=λDd

ω=(4737×1010m)(1.33m)(1×103m)=6.3×104mω=0.63mm

(b) Let t be the thickness of the glass slab. Path difference due to glass slab at centre O.

Δx=μglass μliquid 1t=1.531.331t

 or Δx=0.15t

Now, for the intensity to be minimum at sO, this path difference should be equal to λ2

Δx=λ2 or 0.15t=47372\AAt=15790\AA or t=1.579μm



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