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JEE Advanced 2025 Paper 2 · Physics · Ray and Wave Optics · medium

In a Young's double-slit experiment in air, a thin transparent sheet of refractive index 1.5 and uniform thickness t is placed in front of one slit.…

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In a Young's double-slit experiment in air, a thin transparent sheet of refractive index 1.51.5 and uniform thickness tt is placed in front of one slit. The central bright fringe shifts by exactly 1010 fringe widths. The entire apparatus (slits, sheet and screen, with the sheet still covering the same slit) is now immersed in water of refractive index 4/34/3. By how many fringe widths (in water) does the central fringe now lie displaced from the position it would occupy with no sheet?
  1. A. 103\dfrac{10}{3}
  2. B. 55
  3. C. 1010
  4. D. 203\dfrac{20}{3}
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Answer: (A) 103\dfrac{10}{3}

In a medium of index nmn_m the wavelength is λ/nm\lambda/n_m and the optical path difference introduced by the sheet is (ns−nm)t(n_s-n_m)t measured in vacuum-equivalent length, where ns=1.5n_s=1.5. The fringe width corresponds to a path difference of one wavelength in vacuum units, λ\lambda, so the shift in fringe widths is N=(ns−nm)t/λN=(n_s-n_m)t/\lambda. In air (nm=1n_m=1): 0.5 t=10λ0.5\,t=10\lambda, so t=20λt=20\lambda. In water (nm=4/3n_m=4/3): N=(1.5−43)×20=16×20=103N=(1.5-\tfrac43)\times20=\tfrac16\times20=\tfrac{10}{3}. The shift therefore decreases to 103≈3.33\tfrac{10}{3}\approx3.33 fringe widths. Choosing 1010 ignores the medium; choosing 55 or 20/320/3 comes from using ns/nm−1n_s/n_m-1 without converting fringe width, or other wrong ratios.
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