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 Young's 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. \[\therefore \] \[\] \[{{X}_{30}}=30\beta \] \[{{X}_{20}}=20\beta \] \[\therefore \] \[\frac{{{X}_{30}}}{{{X}_{20}}}=\frac{{{t}_{1}}}{{{t}_{2}}}\] \[\frac{30\beta }{20\beta }=\frac{4.8}{{{t}_{2}}}\] \[{{t}_{2}}=3.2\,mm.\]You need to login to perform this action.
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