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14.   In Young’s experiment, the upper slit is covered by a thin glass plate of refractive index 1.4 while the lower slit is covered by another glass plate, having the same thickness as the first one but having refractive index 1.7. Interference pattern is observed using light of wavelength 5400 Å. It is found that the point P on the screen where the central maximum (n = 0) fell before the glass plates were inserted now has 3/4 the original intensity. It is further observed that what used to be the fifth maximum earlier, lies below the point P while the sixth minimum lies above P. Calculate the thickness of the glass plate. (Absorption
one year ago

From the given data ,μ1 = 1.4 and μ2 = 1.7 . Let  t be the thickness of each glass plates. Path difference at O, due to insertion of glass plates will be

Δx =  ( μ2– μ1) t =  (1.7 – 1.4) t =  0.3 t    …(1) . The point to deduce here is that the path dfference would be lying between 5 λ  and 5λ + λ/2 , since the 5th maxima and the 6th minima lies on both sides of O .

Hence , Δx is taken as    5λ + Δ     …(2) . Here ,   Δ

Φ  =  2π /λ . Δx    =    2π /λ  ( 5λ + Δ )    =  ( 10 π + 2π/λ   Δ )   …… (3)

Now , we know that I (Φ ) = Imax cos²  [ Φ / 2 ]   and the intensity at O is given as 3/4 I max . Therefore ,

3/4 I max =  I (Φ ) = Imax cos²  [ Φ / 2 ]       or    3/4  =  cos²  [ Φ / 2 ]   ….. (4)

Substituting the value of eq (3 ) in eq. (4) and solving , we get  Δ = λ /6 .

Hence , Δx =  5λ + λ /6  = 31 λ / 6  .   Now from eq. (1)  Δx =  0.3 t  .  Therefore  ,

t =  31 λ/6  x  0.3   =   9.3 x 10-6 m = 9.3 microohn
one year ago
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