In a Young’s double slit experiment, the path difference, at a ce… - Vidyadip

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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