Two coherent plane electromagnetic waves of wavelength $0.5 \ \mu\text{m}$ (both have the same amplitude and are linearly polarized along the $z$-direction) fall on the $y = 0$ plane. Their wave vectors $\mathbf{k}_1$ and $\mathbf{k}_2$ are as shown in the figure. If the angle $\theta$ is $30^\circ$, the fringe spacing of the interference pattern produced on the plane is
Let's solve the problem step-by-step to find the fringe spacing of the interference pattern produced by the two coherent plane electromagnetic waves.
Given:
- Wavelength of the waves, \(\lambda = 0.5 \ \mu\text{m}\).
- Angle between each wave vector and the normal to the plane, \(\theta = 30^\circ\).
The fringe spacing \(d\) in the interference pattern can be calculated using the formula:
\(d = \frac{\lambda}{2 \sin\theta}\)
Substituting the given values:
\(d = \frac{0.5 \ \mu\text{m}}{2 \sin 30^\circ}\)
Since \(\sin 30^\circ = 0.5\), we have:
\(d = \frac{0.5 \ \mu\text{m}}{2 \times 0.5} = \frac{0.5 \ \mu\text{m}}{1} = 0.5 \ \mu\text{m}\)
Therefore, the fringe spacing is \(0.5 \ \mu\text{m}\).
Thus, the correct answer is:
- \(0.5 \ \mu\text{m}\)
Important Questions from Optics and Diffraction
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If the wavelength of the light is $\lambda$, the value of $d$ for which the screen would be dark is
