In a single-slit diffraction pattern,if the source of light is replaced by a source of shorter wavelength,the width of the central maximum will:

$A$ parallel beam of monochromatic light is incident on a narrow rectangular slit of width $1\,mm$. When the diffraction pattern is seen on a screen placed at a distance of $2\,m$, the width of the principal maxima is found to be $2.5\,mm$. The wavelength of light is $.............\mathring{A}$.

The distance between the first and the sixth minima in the diffraction pattern of a single slit is $0.5 \, mm$. The screen is $0.5 \, m$ away from the slit. If the wavelength of light used is $5000 \, \mathring{A}$,then the slit width will be ....... $mm$.

The diffraction fringes obtained by a single slit are of

Visible light of wavelength $6000 \times 10^{-8} \; cm$ falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at $60^{\circ}$ from the central maximum. If the first minimum is produced at $\theta_{1}$,then $\theta_{1}$ is close to.....$^{\circ}$

$A$ 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 $Å$)