Question

Mention the required conditions for the interference of light. In Young's double slit experiment, the intensity of light at a point on the screen is \( I \), when path difference is \( \lambda \) by using monochromatic light of wavelength \( \lambda \). Find the intensity of light at that point, where path difference is \( \frac{\lambda}{4} \).

Hint:

In interference, the intensity depends on the phase difference between the two waves. When the path difference is \( \lambda \), the waves are in phase and constructive interference occurs, resulting in maximum intensity. For a path difference of \( \frac{\lambda}{4} \), the waves are out of phase by \( 90^\circ \), leading to partial destructive interference.

Answer 1

Conditions for Interference:
The main conditions for the interference of light are as follows: 1. Coherence: The two light sources must be coherent, meaning they must have a constant phase relationship. 2. Monochromatic light: The light used should be monochromatic (of a single wavelength), as interference depends on the wavelength of the light. 3. Same amplitude: The amplitudes of the two interfering waves should be the same or nearly the same. 4. Superposition principle: The resultant intensity at any point is the sum of the individual intensities due to each wave at that point, following the principle of superposition. Intensity of Light when Path Difference is \( \frac{\lambda}{4} \):
In Young's double-slit experiment, the intensity of light at a point on the screen is given by the equation: \[ I = I_{\text{max}} \cos^2 \left( \frac{\pi \Delta x}{\lambda} \right), \] where: - \( I \) is the intensity at a given point,
- \( I_{\text{max}} \) is the maximum intensity,
- \( \Delta x \) is the path difference between the two waves,
- \( \lambda \) is the wavelength of the monochromatic light. For path difference \( \Delta x = \frac{\lambda}{4} \), we substitute into the equation: \[ I = I_{\text{max}} \cos^2 \left( \frac{\pi \cdot \frac{\lambda}{4}}{\lambda} \right) = I_{\text{max}} \cos^2 \left( \frac{\pi}{4} \right). \] Since \( \cos \left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}} \), we have: \[ I = I_{\text{max}} \left( \frac{1}{\sqrt{2}} \right)^2 = \frac{I_{\text{max}}}{2}. \] Thus, the intensity of light at a point where the path difference is \( \frac{\lambda}{4} \) is half of the maximum intensity: \[ I = \frac{I_{\text{max}}}{2}. \]