Light is incident normally on a diffraction grating through which the first order diffraction is seen at $32^\circ$. The second order diffraction will be seen at

In a single slit diffraction pattern,a light of wavelength $6000 \mathring A$ is used. The distance between the first and third minima in the diffraction pattern is found to be $3 \text{ mm}$ when the screen is placed $50 \text{ cm}$ away from the slit. The width of the slit is . . . . . . $\times 10^{-4} \text{ m}$.

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.

In a single slit diffraction experiment, the first minimum for red light $(660 \ nm)$ coincides with the first secondary maximum of some other wavelength $\lambda$. The value of $\lambda$ is $..... \mathring{A}$.

In a Fraunhofer diffraction at a single slit of width $d$ and incident light of wavelength $5500 \text{ Å}$,the first minimum is observed at an angle $30^{\circ}$. The first secondary maxima are observed at an angle $\theta=$

$A$ circular disc is placed in front of a narrow source. When the point of observation is at a distance of $1 \, m$ from the disc,the disc covers the first $HPZ$ (Half Period Zone). The intensity at this point is $I_0$. The intensity at a point at a distance of $25 \, cm$ from the disc will be: