A) 3.8 mm
B) 1.6 mm
C) 7.6 mm
D) 3.2 mm
Correct Answer: D
Solution :
If shift is equivalent to n fringes then \[n=\frac{(\mu -1)t}{\lambda }\Rightarrow n\propto t\Rightarrow \frac{{{t}_{2}}}{{{t}_{1}}}=\frac{{{n}_{2}}}{{{n}_{1}}}\Rightarrow {{t}_{2}}=\frac{{{n}_{2}}}{{{n}_{1}}}\times t\] \[{{t}_{2}}=\frac{20}{30}\times 4.8=3.2\ mm.\]
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