A) 4 mm
B) 5.6 mm
C) 14 mm
D) 28 mm
Correct Answer: D
Solution :
Let nth minima of 400 nm coincides with mth minima of 560 nm then \[(2n-1)400=(2m-1)560\] Þ \[\frac{2n-1}{2m-1}=\frac{7}{5}=\frac{14}{10}=\frac{21}{15}\] i.e. 4th minima of 400 nm coincides with 3rd minima of 560 nm. The location of this minima is \[=\frac{7(1000)(400\times {{10}^{-6}})}{2\times 0.1}=14\,mm\] Next, 11th minima of 400 nm will coincide with 8th minima of 560 nm Location of this minima is \[=\frac{21(1000)(400\times {{10}^{-6}})}{2\times 0.1}=42\,mm\] \ Required distance = 28 mm
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