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 $\pi = 3.14$)

Light of wavelength $6000 \, \mathring{A}$ is incident on a slit of width $0.30 \, mm$. The screen is placed $2 \, m$ from the slit. Find the position of the first minima.

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 $D$ is equal to:

$A$ diffraction pattern is obtained by making blue light incident on a narrow slit. If blue light is replaced by red light,then:

In an experiment,electrons are made to pass through a narrow slit of width $'d'$ comparable to their de-Broglie wavelength. They are detected on a screen at a distance $'D'$ from the slit (see figure). Which of the following graphs can be expected to represent the number of electrons $'N'$ detected as a function of the detector position $'y'$ ($y = 0$ corresponds to the middle of the slit)?

In a single slit diffraction experiment,for a wavelength of light $\lambda$,the half-angular width of the central maxima is $\theta$. For a wavelength of light $p\lambda$,the half-angular width of the central maxima is $q\theta$. What is the ratio of the half-angular widths of the first secondary maxima in the first case to the second case?