$A$ single slit of width $a$ is illuminated by a monochromatic light of wavelength $600 \, nm$. The value of $a$ for which the first minimum appears at $\theta = 30^{\circ}$ on the screen will be ........... $\mu m$.

$A$ diffraction pattern is obtained using a beam of red light. If the red light is replaced by blue light,then:

$A$ 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 $Å$)

Light of wavelength $600 \,nm$ is incident normally on a slit of width $0.2 \,mm$. The angular width of the central maxima in the diffraction pattern is (measured from minimum to minimum):

$A$ monochromatic light is incident on a single slit of width $0.014 \ mm$. The angular position of the second bright line observed is $2.81^{\circ}$. Then the wavelength of the incident light is $\left[\sin \left(2.81^{\circ}\right)=0.049072\right]$ (in $Å$)

In Fraunhofer diffraction by a single slit,if the slit width is $a$,the wavelength is $\lambda$,and the focal length of the lens is $f$,what is the linear width of the central maximum?