MCQ
In $Y.D.S.E$. using light of wavelength $\lambda$, intensity of light at a point on the screen with path differnce $\lambda$ is $M$ unit. Calculate intensity of light at a point where path difference is $\frac{\lambda }{3}$ is
  • A
    $\frac{M}{2}$
  • $\frac{M}{4}$
  • C
    $\frac{M}{8}$
  • D
    $\frac{M}{16}$

Answer

Correct option: B.
$\frac{M}{4}$
b
$I_{1}=I_{2}=I$

if $\Delta \mathrm{x}=1 \rightarrow$ constructive interference

 then $ \mathrm{I}_{\mathrm{R}} =\mathrm{I}_{\max }=4 \mathrm{I}=\mathrm{M} $

$\mathrm{I}_{1}= \mathrm{I}_{2}=\frac{\mathrm{M}}{4} $

Now, $\Delta x=\frac{\lambda}{3} \Rightarrow \Delta \phi=\frac{2 \pi}{3}$

$\mathrm{I}_{\mathrm{R}}=\mathrm{I}_{1}+\mathrm{I}_{2}+2 \sqrt{\mathrm{I}_{1}} \cdot \sqrt{\mathrm{I}_{2}} \cos \Delta \phi$

$=\frac{M}{4}+\frac{M}{4}+2 \sqrt{\frac{M}{4}} \sqrt{\frac{M}{4}} \cos \frac{2 \pi}{3}$

$\left(\because \cos \frac{2 \pi}{3}=-\frac{1}{2}\right)$

$I_{R}=\frac{M}{4}$

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