A) 0.2 mm
B) 0.4 mm
C) 0.5mm
D) 0.6 mm
Correct Answer: C
Solution :
Key Idea: Refractive index \[\mu =\frac{{{\lambda }_{a}}}{{{\lambda }_{w}}}.\] The fringe width (W) is given by \[W=\frac{D\lambda }{d}\] where,\[\lambda \]is wavelength, D is distance between source and screen and d is distance between coherent sources. \[\ \therefore \] \[\frac{{{W}_{a}}}{{{W}_{w}}}=\frac{{{\lambda }_{a}}}{{{\lambda }_{w}}}\] Also, \[\frac{{{\lambda }_{a}}}{{{\lambda }_{w}}}=\mu \] \[\therefore \] \[\frac{0.8}{{{W}_{w}}}=\mu =1.6\] \[\Rightarrow \] \[{{W}_{w}}=\frac{0.8}{1.6}=\frac{1}{2}=0.5\,mm\]
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