A) 1.8
B) 1.5
C) 1.3
D) 1.6
Correct Answer: A
Solution :
[a] Fringe width \[\omega =\frac{\lambda D}{d}\] When the apparatus is immersed in a liquid, then \[\lambda \] will decrease \[\mu \]times and hence\[\omega \]is reduced \[\mu \] (refractive index) times. \[10\omega '=(5.5)\omega \] or \[10\lambda '\left( \frac{D}{d} \right)=(5.5)\frac{\lambda D}{d}\] or \[\frac{\lambda }{\lambda '}=\frac{10}{5.5}=\mu \] or \[\mu =1.8\]
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