A) 1.6 mm
B) 3.8 mm
C) 3.2 mm
D) 7.6 mm
Correct Answer: C
Solution :
t = thickness of sheet introduced between one of the plates of Youngs slits. \[\beta =\] width of one fringe. \[{{X}_{n}}=\] position of central fringe shifted. \[\therefore \] \[{{X}_{n}}=\frac{D}{2d}(\mu -1)t\] Where D = distance between slit and screen. \[\mu =\] refractive index. 2d= slit width. \[\therefore \] \[\] \[{{X}_{30}}=30\beta \] \[{{X}_{20}}=20\beta \] \[\therefore \] \[\frac{{{X}_{30}}}{{{X}_{20}}}=\frac{{{t}_{1}}}{{{t}_{2}}}\] \[\frac{30\beta }{20\beta }=\frac{4.8}{{{t}_{2}}}\] \[{{t}_{2}}=3.2mm.\]You need to login to perform this action.
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