Two slits are $1 \,mm$ apart and the screen is located $1 \,m$ away from the slits. $A$ light of wavelength $500 \,nm$ is used. The width of each slit to obtain $10$ maxima of the double slit pattern within the central maximum of the single slit pattern is $\ldots \ldots \ldots \times 10^{-4} \,m$.

Due to which optical phenomenon do we not see the shadow of an opaque object clearly?

In a single-slit diffraction pattern,the intensity of the central maximum is $I_0$. If the slit width is doubled,what will be the intensity of the central maximum?

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 $Å$)

Two wavelengths $590 \ nm$ and $596 \ nm$ of sodium light are used one after another to study the diffraction taking place at a single slit of aperture $2.4 \ mm$. The distance between the slit and the screen is $2 \ m$. The separation between the positions of the first secondary maximum of the diffraction pattern obtained in the two cases is: