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

Wave Optics PYQ — JEE Main 2019 · 11 Jan · Shift 1 · Q57 with Solution

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Correct answer: B

For two coherent sources of equal intensity, the intensity at path difference $\Delta$ is

I = I_{\max} \cos^2\left(\frac{\phi}{2}\right)

where phase difference

\phi = \frac{2\pi \Delta}{\lambda}

At the centre of a bright fringe, $\Delta = 0$ , so intensity is maximum:

I_{\max}

Thus,

\frac{I}{I_{\max}} = \cos^2\left(\frac{\phi}{2}\right)

\Delta = \frac{\lambda}{8}

So the phase difference is

\phi = \frac{2\pi}{\lambda}\cdot \frac{\lambda}{8} = \frac{\pi}{4}

Hence,

\frac{\phi}{2} = \frac{\pi}{8}

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

Using

\cos\left(\frac{\pi}{8}\right) \approx 0.924

therefore,

\cos^2\left(\frac{\pi}{8}\right) \approx (0.924)^2 \approx 0.854

So,

\frac{I}{I_{\max}} \approx 0.85

The closest option is:

\boxed{0.85}

So the correct option is B.