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  • The path difference between two interfering waves at a point on the screen is

The path difference between two interfering waves at a point on the screen is \mathrm{\lambda / 8}. The ratio of the intensity at this point and that at the central fringe will be:

Option: 1

0.853


Option: 2

8.53


Option: 3

85.3


Option: 4

853


Answers (1)

best_answer

As  \mathrm{R^2=a^2+b^2+2 a b \cos \phi}

and path difference of \mathrm{\lambda / 8} is equivalent to phase difference of

\mathrm{\phi=\frac{2 \pi}{8}=\frac{\pi}{4}}

\mathrm{\begin{array}{ll} \therefore & \mathrm{I}_{\mathrm{R}}=\mathrm{I}+\mathrm{I}+2 \sqrt{\mathrm{II}} \cos \frac{\pi}{4}=2 \mathrm{I}\left(1+\frac{1}{\sqrt{2}}\right)=3.414 \mathrm{I} \\\\ & \mathrm{I}_{\max }=4 \mathrm{I} \\\\ \therefore & \frac{\mathrm{I}_{\mathrm{R}}}{\mathrm{I}_{\max }}=\frac{3.414 \mathrm{I}}{4 \mathrm{I}}=0.853 \end{array}}

Posted by

Ajit Kumar Dubey

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