Question

In a Young's double slit experiment, the path difference, at a certain point on the screen, between two interfering waves is ${\frac{1}{8}}^{th}$ of wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to:

• A. $0.74$
• B. $0.85$
• C. $0.94$
• D. $0.80$

Answer 1

The correct option is B $0.85$

Given that,
$\text{path difference,}\Delta x=\frac{\lambda }{8}$

Phase difference $\left(\Delta \varphi \right)$ is given by
$\Delta \varphi =\frac{2\pi }{\lambda }\left(\Delta x\right)$

$\Delta \varphi =\frac{\left(2\pi \right)}{\lambda }\frac{\lambda }{8}=\frac{\pi }{4}$

For two sources with same intensity, in different phases,

$I=I_{max}{\cos }^2\left(\frac{\Delta \varphi }{2}\right)$

$I=I_{max}{\cos }^2\left(\frac{\pi }{8}\right)$

The intensity at the centre of the bright fringe is given by,

$I_0=I_{max}$

$\Rightarrow \frac{I}{I_0}={\cos }^2\left(\frac{\pi }{8}\right)$

$\Rightarrow \frac{I}{I_0}=\frac{1+\cos \frac{\pi }{4}}{2}=\frac{1+\frac{1}{\sqrt{2}}}{2}$

$\Rightarrow \frac{I}{I_0}=0.85$

Hence, option $\left(B\right)$ is correct.