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

Gujarat BoardEnglish MediumSTD 11 ScienceNEETSTD 12 - 10. Wave optics4 Marks
MCQ
In a Young’s double slit experiment, the path difference, at a certain point on the screen, between two interfering waves is $1/8^{th}$ of wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to
  • A
    $0.74$
  • ✓
    $0.85$
  • C
    $0.94$
  • D
    $0.80$
✓

Answer

Correct option: B.
$0.85$
b
$\Delta x=\frac{\lambda}{8}$

Phase $|\Delta P|=\frac{2 \pi}{\lambda} \times \frac{\lambda}{8}=\frac{\pi}{4}$

$\therefore $ $\mathrm{I}_{\mathrm{res}}=I+I+2I \cos \left(\frac{\mathrm{R}}{4}\right)$

$=\,2I\left(1+\frac{1}{\sqrt{2}}\right)=2I \times 1.7$

$\therefore \quad \frac{I_{res }}{I_{\operatorname{man}}}=\frac{2I\times 1.7}{41}=0.85$

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