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:

$A$ 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:

$A$ slit of width $a$ is illuminated by white light. For red light $(\lambda = 6500 \; \mathring{A})$,the first minima is obtained at $\theta = 30^\circ$. Then the value of $a$ will be

In a diffraction pattern due to a single slit of width $a$,the first minimum is observed at an angle $30^{\circ}$ when light of wavelength $5000 \; \mathring{A}$ is incident on the slit. The first secondary maximum is observed at an angle of

In a single slit diffraction pattern, the distance between the plane of the slit and the screen is $1.4 \text{ m}$. The width of the slit is $0.66 \text{ mm}$. The second maximum is formed at the distance of $2.8 \text{ mm}$ from the center of the screen. The wavelength of light used is (in $\text{ Å}$)

$A$ 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 $Å$)