A) 8 mm
B) 6 mm
C) 4 mm
D) 10 mm
Correct Answer: A
Solution :
Shift \[\Delta x=\frac{\beta }{\lambda }(\mu -1)\,t\] Shift due to one plate \[\Delta {{x}_{1}}=\frac{\beta }{\lambda }({{\mu }_{1}}-1)\] Shift due to another path \[\Delta {{x}_{2}}=\frac{\beta }{\lambda }({{\mu }_{2}}-1)\,t\] Net shift \[\Delta x=\Delta {{x}_{2}}-\Delta {{x}_{1}}=\frac{\beta }{\lambda }({{\mu }_{2}}-{{\mu }_{1}})\,t\] ?..(i) Also it is given that \[\Delta x=5\beta \] ?..(ii) Hence \[5\beta =\frac{\beta }{\lambda }({{\mu }_{1}}-{{\mu }_{2}})\,t\] \[\Rightarrow t=\frac{5\lambda }{({{\mu }_{2}}-{{\mu }_{1}})}=\frac{5\times 4800\times {{10}^{-10}}}{(1.7-1.4)}\]\[=8\times {{10}^{-6}}m=8\mu \,m.\]You need to login to perform this action.
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