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:
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:
Three identical pinholes separated by distance $a$ along the x-axis are illuminated by a collimated monochromatic coherent beam of light (wavelength $\lambda$) as shown in the figure below.

The intensity (in arbitrary units) pattern of fringes obtained on a screen kept at distance $D$ ($D>>a$) along the z- axis is best represented by
The figure below describes the arrangement of slits and screens in a Young's double slit experiment. The width of the slit in $\text{S}_1$ is $a$ and the slits in $\text{S}_2$ are of negligible width.
If the wavelength of the light is $\lambda$, the value of $d$ for which the screen would be dark is
A screen has two slits, each of width $w$, with their centres at a distance $2w$ apart. It is illuminated by a monochromatic plane wave travelling along the $x$-axis.

The intensity of the interference pattern, measured on a distant screen, at an angle $\theta = n\lambda/w$ to the $x$-axis is