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JEE Advanced Physics Wave Optics 2025 JEE Advanced 2025 (Paper 2)

JEE Advanced Physics Question (2025) — Solution

Question

In a Young's double slit experiment, a combination of two glass wedges A and B, having refractive indices 1.7 and 1.5, respectively, are placed in front of the slits, as shown in the figure. The separation between the slits is d=2 ~mm and the shortest distance between the slits and the screen is D=2 ~m . Thickness of the combination of the wedges is t=12 ~m . The value of l as shown in the figure is 1 mm . Neglect any refraction effect at the slanted interface of the wedges. Due to the combination of the wedges, the central maximum shifts (in mm ) with respect to O by ______

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

aligned & x+y=12 m \\ & 4 12 = 1 x \\ & x=3 m \\ & y=6 m aligned aligned & = y d D - ( _B-1 ) x- ( _A-1 ) y+ ( _B-1 ) y+ ( _A-1 ) x \\ & -y d D =-0.5 3-0.7 9+0.5 9+0.7 3 \\ & -y d D =-0.5 6-0.7 6 \\ & -y d D =-1.2 ~m \\ & y= 1.2 D d = 1.2 2 2 10^ -3 10^ -6 \\ & =1.2 ~mm aligned

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