Question:

In a Young's double-slit experiment, two light waves, each of intensity \( I_0 \), interfere at a point, having a path difference \( \frac{\lambda}{8} \) on the screen. Find the intensity at this point.

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In interference patterns, the intensity is determined by the phase difference, which depends on the path difference between the waves.
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Solution and Explanation

The intensity in an interference pattern is given by: \[ I = I_0 \left( 1 + \cos \delta \right) \] where \( \delta \) is the phase difference given by: \[ \delta = \frac{2\pi}{\lambda} \cdot \text{path difference} \] Substitute the given path difference \( \frac{\lambda}{8} \): \[ \delta = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{8} = \frac{\pi}{4} \] Thus, the intensity is: \[ I = I_0 \left( 1 + \cos \frac{\pi}{4} \right) = I_0 \left( 1 + \frac{\sqrt{2}}{2} \right) \] \[ I = I_0 \left( 1 + 0.707 \right) = 1.707 I_0 \] Thus, the intensity at the point is \( 1.707 I_0 \). 

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