MCQ
The path difference between two interfering waves of equal intensities at a point on the screen is $\frac{\lambda }{4}.$ The ratio of intensity at this point and that at the central fringe will be
- A$1 : 1$
- ✓$1 : 2$
- C$2 : 1$
- D$1 : 4$
✓
Answer
Correct option: B.
$1 : 2$
b
(b) By using $I = 4{I_0}{\cos ^2}\left( {\frac{\phi }{2}} \right) = 4{I_0}{\cos ^2}\left( {\frac{{\pi \Delta }}{\lambda }} \right)$
(b) By using $I = 4{I_0}{\cos ^2}\left( {\frac{\phi }{2}} \right) = 4{I_0}{\cos ^2}\left( {\frac{{\pi \Delta }}{\lambda }} \right)$
$\left\{ {\because \phi = \frac{{2\pi }}{\lambda }\Delta } \right\}$
==> $\frac{{{I_1}}}{{{I_2}}} = \frac{{{{\cos }^2}\left( {\frac{{\pi {\Delta _1}}}{\lambda }} \right)}}{{{{\cos }^2}\left( {\frac{{\pi {\Delta _2}}}{\lambda }} \right)}} = \frac{{{{\cos }^2}\left( {\frac{{\pi .\frac{\lambda }{4}}}{\lambda }} \right)}}{{{{\cos }^2}(0)}} = \frac{1}{2}$
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