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 . 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 [figure]
Answer
1.2
Solution
Height of the wedge block =d+2l=4 mm; thickness of A grows linearly from 0 at the bottom to t at the top.
Upper slit (3 mm from bottom): tA= 34t, tB= 14t; lower slit: tA= 14t, tB= 34t.
Optical paths: upper =1.7 ( 34t )+1.5 ( 14t )=1.65t; lower =1.7 ( 14t )+1.5 ( 34t )=1.55t
Path difference =0.1t=1.2
Shift = Dd= 1.2 10-6 22 10-3=1.2 10-3 m =1.2 mm (towards the upper slit)
Answer: 1.2