Light of wavelength $6000 \, \mathring A$ is incident on a slit of width $0.30 \, mm$. $A$ screen is placed at a distance of $2 \, m$ from the slit. Find the position of the first minimum.

$A$ single slit diffraction pattern is formed with light of wavelength $6384 Å$. The second secondary maximum for this wavelength coincides with the third secondary maximum in the pattern for light of wavelength $\lambda_0$. The value of $\lambda_0$ is (in $Å$)

In a single slit diffraction experiment, the width of the slit is reduced. What happens to the linear width of the principal maxima?

Two slits are $1 \,mm$ apart and the screen is located $1 \,m$ away from the slits. $A$ light of wavelength $500 \,nm$ is used. The width of each slit to obtain $10$ maxima of the double slit pattern within the central maximum of the single slit pattern is $\ldots \ldots \ldots \times 10^{-4} \,m$.

$A$ slit of width $a$ is illuminated by monochromatic light of wavelength $650 \ nm$. The value of $a$ when the first maximum is formed at a diffraction angle of $30^\circ$ is:

Light of wavelength $\lambda$ is incident on a slit of width $a$. The resulting diffraction pattern is observed on a screen placed at a distance $D$. If the linear width of the central maximum is equal to the width of the slit,then $D = $ . . . . . . .