Kumar Mittal Solutions for Exercise 2: For DIFFERENT COMPETITIVE EXAMINATIONS

A parallel monochromatic beam of light is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the first minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of the slit is:

A beam of light of $\lambda =600 \mathrm{nm}$ from a distant source falls on a single-slit $1 \mathrm{mm}$ wide and the resulting diffraction pattern is observed on a screen $2 \mathrm{m}$ away. The distance between first dark fringes on either side of the central bright fringe is:

For a parallel beam of monochromatic light of wavelength $\lambda$, diffraction is produced by a single slit whose width $e$ is of the wavelength of light. If $D$ is the distance of the screen from the slit, the width of the central maxima will be:

In a double-slit experiment, the two slits are $1 \mathrm{mm}$ apart and the screen is placed $1 \mathrm{m}$ away. A monochromatic light of wavelength $500 \mathrm{nm}$ is used. What will be the width of each slit for obtaining ten maxima of double-slit within the central maxima of single-slit pattern?

A linear aperture whose width is $0.02 \mathrm{cm}$ is placed immediately in front of a lens of focal length $60 \mathrm{cm}$. The aperture is illuminated normally by a parallel beam of light of wavelength $5\times {10}^{-5} \mathrm{cm}$. The distance of the first dark band of the diffraction pattern from the centre of the screen is:

In a diffraction pattern due to a single slit of width $e$, the first minimum is observed at an angle $30°$ when light of wavelength $5000 \stackrel{\mathrm{o}}{\mathrm{A}}$ is incident on the slit. The first secondary is observed at an angle of:

The box of a pin hole camera of length $L$ has a hole of radius $e$. It is assumed that when the hole is illuminated by a parallel beam of light of wavelength $\lambda$, the spread of the spot (obtained on the opposite wall of the camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would then have its minimum size (say $b_{\min }$) when:

The angular width of the central maximum in a single slit diffraction pattern is $60°$. The width of the slit is $1 \mathrm{\mu m}$. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young’s fringes can be observed on a screen placed at a distance $50 \mathrm{cm}$ from the slits. If the observed fringe width is $1 \mathrm{cm}$, what is slit separation distance? (i.e., distance between the centres of each slit.)