A) 90
B) 12
C) 14
D) 24
Correct Answer: D
Solution :
[d] Shift of fringe pattern \[=(\mu -1)\frac{tD}{d}\] \[\therefore \,\,\,\frac{30D(4800\times {{10}^{-10}})}{d}=(0.6)t\frac{D}{d}\] \[30\times 4800\times {{10}^{-10}}=0.6t\] \[t=\frac{30\times 4800\times {{10}^{-10}}}{0.6}=\frac{1.44\times {{10}^{-5}}}{0.6}=24\times {{10}^{-6}}\]
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