$A$ screen is placed $50 \, cm$ from a single slit which is illuminated with $6000 \, \mathring{A}$. If the distance between the first and third minima in the diffraction pattern is $3 \, mm$,what is the width of the slit?

In a single slit diffraction pattern,the wavelength of light used is $628 \text{ nm}$ and slit width is $0.2 \text{ mm}$. The angular width of the central maximum is $\alpha \times 10^{-2} \text{ degrees}$. The value of $\alpha$ is . . . . . . . (Take $\pi = 3.14$)

$A$ single slit diffraction pattern is formed with light of wavelength $6195 Å$. 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 $\lambda = 5000 \ \mathring{A}$ 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 ....... $mm$.

$A$ parallel beam of light of wavelength $\lambda$ is incident normally on a single slit of width $d$. Diffraction bands are obtained on a screen placed at a distance $D$ from the slit. The second dark band from the central bright band will be at a distance given by

$A$ diffraction pattern is obtained by using a beam of red light. What will happen if the red light is replaced by blue light?