Question

Differentiate between interference and diffraction of light.

Two coherent light waves, each of intensity $I_0$ produce interference pattern on a screen. Find the intensity at a point at which the path difference between them is (i) $\dfrac{\lambda}{3}$, and (ii) $\dfrac{\lambda}{4}$.

Hint:

Use \(I = 4I_0\cos^2(\phi/2)\) with \(\phi = 2\pi\Delta x/\lambda\).

Answer 1

Step 1: Understanding the Concept:
Interference is the redistribution of energy when two coherent waves overlap. Diffraction is the bending of light around the edge of an obstacle or slit.
For two waves of equal intensity \(I_0\), the resultant intensity is \(I = 4I_0\cos^2(\phi/2)\), where \(\phi\) is the phase difference.

Step 2: Difference between interference and diffraction:
1. Interference comes from the superposition of waves from two coherent sources. Diffraction comes from different parts of one wavefront.
2. In interference, all bright fringes have the same intensity. In diffraction, the intensity falls quickly for higher-order maxima.
3. Interference fringes are usually of equal width. Diffraction fringes (central maximum) are unequal in width: the central one is twice as wide as the others.

Step 3: Intensity for path difference \(\lambda/3\):
\[ \phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{3} = \frac{2\pi}{3} \]
\[ I = 4I_0\cos^2\left(\frac{\pi}{3}\right) = 4I_0 \times \frac{1}{4} = I_0 \]

Step 4: Intensity for path difference \(\lambda/4\):
\[ \phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2} \]
\[ I = 4I_0\cos^2\left(\frac{\pi}{4}\right) = 4I_0 \times \frac{1}{2} = 2I_0 \]

Final Answer:
The intensity is \(I_0\) for a path difference \(\lambda/3\) and \(2I_0\) for \(\lambda/4\). \[ \boxed{I_{(i)} = I_0,\quad I_{(ii)} = 2I_0} \]