A) 1.5
B) 1.2
C) 1.3
D) 1.7
Correct Answer: A
Solution :
The position of 30th bright fringe \[{{y}_{30}}=\frac{30\lambda D}{d}\] Now, position shift to central fringe is \[{{y}_{0}}=\frac{30\lambda D}{d}\] But we know \[{{y}_{0}}=\frac{D}{d}\,(\mu -1)\,t\] \[\frac{30\lambda D}{d}=\frac{D}{d}\,(\mu -1)\,t\] \[(\mu -1)=\frac{30\lambda }{t}\] \[=\frac{30\times 6000\times {{10}^{-10}}}{3.6\times {{10}^{-5}}}=0.5\] \[\mu =1.5\]You need to login to perform this action.
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