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BITSAT2015PhysicsWave OpticsActual

In a YDSE, the light of wavelength =5000 Ã is used, which emerges in phase from two slits a distance d =3 10⁻⁷ ~m apart. Atransparent sheet of thickness t =1.5 10⁻⁷ ~m refractive index =1.17 is placed over one of the slits. what is the new angular position of the central maxima of the interference pattern, from the centre of the screen? Find the value of y .

Options

  1. A4.9^ and D( -1) t 2 d
  2. B4.9^ and D ( -1) t d
  3. C3.9^ and D( +1) t d
  4. D2.9^ and 2 D ( +1) t d

Correct answer

B. 4.9^ and D ( -1) t d

Step-by-step solution

The path difference when transparent sheet is introduced x=( -1) t If the central maxima occupies position of nth fringe, then ( -1) t = n = d = ( -1) t d = (1.17-1) 1.5 10⁻⁷ 3 10⁻⁷ =0.085 Therefore, angular position of central maxima = ⁻¹(0.085)=4.88^ 4.9 For small angles, = = = y D y D = ( -1) t d y = D ( -1) t d

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