The half angular width of the central maxima in the Fraunhofer diffraction due to a slit of width $\frac{1200}{\sqrt{2}} \ \mu m$ is $45^o$. Then the wavelength of the light is ...... $\mu m$.

Maximum diffraction takes place in a given slit for

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 $6384 Å$. 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 $\lambda$ is incident on a single slit of width $a$,and the distance between the slit and the screen is $D$. In the diffraction pattern,if the slit width is equal to the width of the central maximum,then $D=$

$A$ slit of size $0.15 \,cm$ is placed at $2.1 \,m$ from a screen. On illuminating it by a light of wavelength $5 \times 10^{-5} \,cm$,the width of the central maxima will be: