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

Wave Optics PYQ — JEE Main 2020 · 8 Jan · Shift 2 · Q54 with Solution

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Correct answer: A

For two coherent waves of equal intensity, the intensity at phase difference $\phi$ is

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

where $I_{\max}$ is the intensity at the centre of a bright fringe.

Given path difference

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

The corresponding phase difference is

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

We need

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

So,

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

Now,

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

Hence,

\cos^2\left(\frac{\pi}{8}\right) \approx (0.924)^2 \approx 0.853

\frac{I}{I_{\max}} \approx 0.853

So the correct option is A.

Stored correct answer: A

Our derived answer: A

They agree.