Red light of wavelength $5400 \ \mathring{A}$ from a distant source falls on a slit $0.80 \ mm$ wide. Calculate the distance between the first two dark bands on each side of the central bright band in the diffraction pattern observed on a screen placed $1.4 \ m$ from the slit. (in $mm$)

Light of wavelength $5000 \, \mathring{A}$ is incident on a single slit such that the first minima is formed at a distance $5 \, mm$ from the centre. If the screen is placed $2 \, m$ away,then find the width of the slit in $mm$.

In the case of Fraunhofer diffraction due to a single slit,the intensity of the central maximum is $I_0$. If the width of the slit is doubled,what will be the new intensity of the central maximum?

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 = $ . . . . . . .

In a single slit diffraction, the slit is illuminated with light of wavelength $6000 \text{ Å}$. If the slit were illuminated by a light of wavelength $\lambda$ the angular width of the central maximum decreases by $30 \%$. Then $\lambda$ is (in $\text{ Å}$)

In a single slit diffraction pattern,a light of wavelength $6000 \mathring A$ is used. The distance between the first and third minima in the diffraction pattern is found to be $3 \text{ mm}$ when the screen is placed $50 \text{ cm}$ away from the slit. The width of the slit is . . . . . . $\times 10^{-4} \text{ m}$.