The ratio of intensities at two points P and Q on the screen in a Young's double slit experiment where phase difference

Question

The ratio of intensities at two points P and Q on the screen in a Young's double slit experiment where phase difference between two waves of same amplitude are 3 and 2, respectively, are:

Accepted Answer

Given: In Young's double slit experiment, the phase differences at points $P$ and $Q$ are $\frac{\pi}{3}$ and $\frac{\pi}{2}$ respectively.

Find: The ratio of intensities at $P$ and $Q$ .

Using the formula for intensity in Young's double slit experiment:

I_{\text{res}} = 4I_0 \cos^2\left(\frac{\theta}{2}\right)

where $I_0$ is the maximum intensity and $\theta$ is the phase difference.

For the first point $P$ , where $\theta = \frac{\pi}{3}$ :

I_1 = 4I_0 \cos^2\left(\frac{\pi}{6}\right)

I_1 = 4I_0 \left(\frac{\sqrt{3}}{2}\right)^2 = 4I_0 \times \frac{3}{4} = 3I_0

For the second point $Q$ , where $\theta = \frac{\pi}{2}$ :

I_2 = 4I_0 \cos^2\left(\frac{\pi}{4}\right)

I_2 = 4I_0 \left(\frac{1}{\sqrt{2}}\right)^2 = 4I_0 \times \frac{1}{2} = 2I_0

Thus, the ratio of intensities is:

\frac{I_1}{I_2} = \frac{3I_0}{2I_0} = \frac{3}{2}

Therefore, the ratio is $3:2$ and the correct option is A.

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