$A$ 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 deriving the single slit diffraction pattern,it is stated that the intensity is zero at angles of $\theta = n\lambda / a$,where $a$ is the slit width. Justify this by suitably dividing the slit to bring out the cancellation.

Light of wavelength $\lambda = 5000 \ \mathring{A}$ falls normally on a narrow slit. $A$ screen is placed at a distance of $1 \ m$ from the slit and perpendicular to the direction of light. The first minima of the diffraction pattern is situated at $5 \ mm$ from the centre of the central maximum. The width of the slit is ....... $mm$.

$A$ light wave of wavelength $\lambda$ is incident on a slit of width $d$. The resulting diffraction pattern is observed on a screen at a distance $D$. If the linear width of the principal maxima is equal to the width of the slit,then the distance $D$ is:

$A$ beam of light of $\lambda = 600 \, nm$ from a distant source falls on a single slit $1 \, mm$ wide and the resulting diffraction pattern is observed on a screen $2 \, m$ away. The distance between the first dark fringes on either side of the central bright fringe is:

In a Fraunhofer diffraction at a single slit,if yellow light illuminating the slit is replaced by blue light,then the diffraction bands