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STD 12 - 10. Wave optics — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 10. Wave optics4 Marks
MCQ
In a Young's double slit experiment the intensity at a point where the path difference is $\frac \lambda {6}$ ($\lambda$  being the wavelength of the light used) is $I.$ If $I_0$ denotes the maximum intensity, $I_0$/$I$ is equal to
  • A
    $\sqrt 2$
  • ✓
    $\frac{4}{3}$
  • C
    $2$
  • D
    $\frac{2}{{\sqrt 3 }}$
✓

Answer

Correct option: B.
$\frac{4}{3}$
b
$I=I_{\max } \cos ^{2} \frac{\phi}{2}$

$\phi \Rightarrow$ Phase difference

When path difference is $\frac{\lambda}{6}$ then

$\phi=\frac{2 \pi 3}{\lambda} \times \frac{\lambda}{6}=\pi$

$I=I_{\max } \cos ^{2} \frac{\pi}{6}$

$I=I_{\max } \times \frac{3}{4}$

$\frac{I_{\max }}{I}=\frac{4}{3}$

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