StreakPeaked· Practice

ExamsJEE AdvancedPhysics

In a Young's double slit experiment (YDSE) performed in a medium of refractive index 4/3, light of wavelength 600 nm in vacuum falls on slits separated by 0.45 mm. The lower slit S2 is covered by a thin glass sheet of thickness 10.4 micrometers and refractive index 1.5. The screen is placed 1.5 m from the slits. Find the location of the central maximum (in mm) on the y-axis from the central point O.

  1. 1
  2. 2
  3. 3
  4. 4

Correct answer: 4

Solution

Extra optical path (in medium) due to sheet = t*(mu_glass - mu_medium) = 10.4e-6*(1.5 - 4/3) = 10.4e-6*(1/6) m. Wait: extra path in vacuum equivalent = t*(mu_glass/mu_medium - 1)... shift y = D*t*(mu_glass - mu_medium)/(d * lambda_medium)... Let me use: extra number of wavelengths = t*(mu_glass - mu_medium)/lambda_vacuum = 10.4e-6*(1.5-4/3)/600e-9 = 10.4e-6*(1/6)/600e-9 = (10.4/6)/0.6 = 1.733/0.6 approx 2.89 fringes. Fringe width in medium = lambda_vacuum*D/(n_medium*d) = 600e-9*1.5/(4/3 * 0.45e-3) = 900e-9/(0.6e-3) = 1.5e-3 m = 1.5 mm. Shift = 2.89 * 1.5 mm = 4.33 mm ≈ 4 mm.

Related JEE Advanced Physics questions

⚔️ Practice JEE Advanced Physics free + battle 1v1 →