The diameters of the pinholes of two otherwise identical cameras $A$ and $B$ are $500$ $\mu \text{m}$ and $200 \text{ } \mu \text{m}$, respectively. Then the image in camera $A$ will be
The sharpness of an image formed by a pinhole camera depends on the pinhole size. A smaller pinhole reduces the spread of light rays, leading to a sharper image. Conversely, a larger pinhole allows more light rays but also increases diffraction effects and the blurring of the image.
Since $D_A > D_B$, Camera A has a larger pinhole, resulting in an image that is less sharp than the image in Camera B.
The brightness of the image is determined by the amount of light that passes through the pinhole. The amount of light is proportional to the area of the pinhole.
The area of a circular pinhole is given by $A = \pi r^2 = \pi (D/2)^2$, which is proportional to the square of the diameter ($D^2$).
Since $A_A > A_B$, Camera A allows more light to enter than Camera B. Therefore, the image in Camera A will be brighter than the image in Camera B.
Combining the effects on sharpness and brightness:
Thus, the image in camera A will be less sharp and brighter than in B.
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
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
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