Yellow light is used in a single slit diffraction experiment with a slit width of $0.6 \, mm$. If the yellow light is replaced by $X$-rays,then the pattern will reveal:

In a single-slit diffraction experiment,the first minimum for a wavelength $\lambda_1 = 660 \ nm$ coincides with the first maximum for a wavelength $\lambda_2$. Find the value of $\lambda_2$ in $nm$.

If the width of the slit is $a$,then the value of the first secondary maximum in a single slit diffraction pattern is given by:

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 $\text{ Å}$)

$A$ Fraunhofer diffraction pattern due to a single slit of width $0.3 \text{ mm}$ is obtained on a screen placed at a distance of $3 \text{ m}$ from the slit. The first minima lie at $5.5 \text{ mm}$ on either side of the central maximum on the screen. The wavelength of light used is (in $\text{ Å}$)

Light of wavelength $\lambda$ is incident on a single slit of width $a$,and the distance between the slit and the screen is $D$. In the diffraction pattern,if the slit width is equal to the width of the central maximum,then $D=$