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Light of wavelength \lambda _{0} in air enters a medium of refractive index n. If two points A and B in this medium lie along the path of this light at a distance x_{1}  then phase difference \phi _{0}  between these two point is :

Option: 1

\phi_{0} =\frac{1}{n}\cdot \left ( \frac{2\pi }{\lambda _{0}} \right )\cdot x


Option: 2

0


Option: 3

\phi_{0} =n\cdot \left ( \frac{2\pi}{\lambda_{0}} \right )\cdot x


Option: 4

\phi_{0} =\frac{1}{n-1}\cdot \left ( \frac{2\pi}{\lambda_{0}} \right )\cdot x


Answers (1)

best_answer

As we learn

Relation between phase & path difference -

\Delta \phi = \frac{2\pi }{\lambda }\times \Delta x

 

- wherein

\Delta \phi = Phase difference

\Delta x= Path Difference

\lambda = Wavelength

  \text{Phase difference}=\frac{2\pi}{\lambda _{0}} \text{(optical path difference)}

Optical path difference = n\cdot x

\phi _{0}=\frac{2\pi }{\lambda _{0}}\cdot \left ( nx \right )

Posted by

Pankaj

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