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.)

This question has Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements.

Statement-1: Out of radio waves and microwaves, the radio waves undergo more diffraction.

Statement-2: Radio waves have greater frequency compared to microwaves.

An observer is moving with half the speed of light towards a stationary microwave source emitting waves at frequency $10 \mathrm{GHz}$. What is the frequency of the microwave measured by the observer? (Speed of light is $3\times {10}^8 \mathrm{m} {\mathrm{s}}^{-1}$).

Two parallel beams of light $P$ and $Q$ (separation $d$) containing radiations of wavelengths $4000 \mathrm{\AA }$ and $5000 \mathrm{\AA }$ (which are mutually coherent in each wavelength separately) are incident normally on a prism as shown in the figure. The refractive index of the prism as a function of wavelength is given by the relation, $\mathrm{\mu } \left(\mathrm{\lambda }\right)=1.20+\frac{b^2}{{\mathrm{\lambda }}^2}$ where $\mathrm{\lambda }$ is in $\mathrm{\AA }$ and $b$ is a positive constant. The value of $b$ is such that the condition for total reflection at the face $AC$ is just satisfied for one wavelength and is not satisfied for the other. A convergent lens is used to bring these transmitted beams into focus. If the intensities of the upper and the lower beams immediately after transmission from the face $AC$, are $4I\mathit{ }$and $I$ respectively, find the resultant intensity at the focus.

White coherent light $(400 \mathrm{nm}-700 \mathrm{nm})$ is sent through the slits of a Young’s double-slit experiment (as shown in the figure). The separation between the slits is $1 \mathrm{mm}$and the screen is $100 \mathrm{cm}$ away from the slits. There is a hole in the screen at a point $1.5 \mathrm{mm}$ away (along the width of the fringes) from the central line. 

$\left(\mathrm{a}\right)$For which wavelength(s) there will be minima at that point?

$\left(\mathrm{b}\right)$ Which wavelength(s) will have a maximum intensity?

A beam of light consisting of two wavelengths, $6500 \mathrm{\AA }$ and $5200 \mathrm{\AA }$ is used in a double slit experiment $(1 \mathrm{\AA } = {10}^{–10} \mathrm{m}).$ The distance between the slits is $2.0 \mathrm{mm}$ and the distance between the plane of the slits and the screen is $120 \mathrm{cm}.$

$\left(\mathrm{a}\right)$ Find the distance of the third bright fringe on the screen from the central maximum for the wavelength $6500 \mathrm{\AA }.$

$\left(\mathrm{b}\right)$ What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?

A source $S$ is kept directly behind the slit $S_1$ in a double-slit apparatus. Find the phase difference at a point $O$ which is equidistant from $S_1$ and $S_2.$ What will be the phase difference at $P$ if a liquid of refraction index $\mathrm{\mu }$ is filled; (wavelength of light in air is $\lambda$ due to the source). Assume same intensity due to $S_1$ and $S_2$  on screen and position at the liquid.$( \mathrm{\lambda }<< d, d<< D, l >> d)$ 

$\left(\mathrm{a}\right)$ between the screen and the slits.

$\left(\mathrm{b}\right)$ between the slits and the source $S$. In this case find the minimum distance between the points on the screen where the intensity is half the maximum intensity on the screen.