I E Irodov Solutions for Chapter: OPTICS, Exercise 2: INTEREFENCE OF LIGHT

Two coherent plane light waves propagating with a divergence angle $\mathrm{\psi }\ll 1$ fall almost normally on a screen. The amplitudes of the waves are equal. Demonstrate that the distance between the neighbouring maxima on the screen is equal to $\mathrm{\Delta }x=\frac{\lambda }{\mathrm{\psi }}$, where $\lambda$ is the wavelength.

Figure illustrates the interference experiment with Fresnel mirrors. The angle between the mirrors is $\alpha ={12}^{\text{'}},$ the distances from the mirrors' intersection line to the narrow slit $S$ and the screen $Sc$ are equal to $r=10.0 \mathrm{cm}$ and $b=130 \mathrm{cm}$, respectively. The wavelength of light is $\lambda =0.55 \mathrm{\mu m}$. Find:
(a) the width of a fringe on the screen and the number of possible maxima;
(b) the shift of the interference pattern on the screen, when the slit is displaced by $\delta l=1.0 \mathrm{mm}$ along the arc of radius $r$, with centre at the point $O$;
(c) at what maximum width ${\delta }_{\max }$ of the slit, the interference fringes on the screen are still observed sufficiently sharp.

A plane light wave falls on Fresnel mirrors with an angle $\alpha =2.0^{\text{'}}$ between them. Determine the wavelength of light, if the width of the fringe on the screen $\Delta x=0.55 \mathrm{mm}$.

A lens of diameter $5.0 \mathrm{cm}$ and focal length $f=25.0 \mathrm{cm}$ was cut along the diameter into two identical halves. In the process, the layer of the lens, $a=1.00 \mathrm{mm}$ in thickness, was lost. Then the halves were put together to form a composite lens. In this focal plane, a narrow slit was placed, emitting monochromatic light with wavelength $\lambda =0.60 \mathrm{\mu m}$. Behind the lens a screen was located at a distance $b=50 \mathrm{cm}$ from it. Find:
(a) the width of a fringe on the screen and the number of possible maxima;
(b) the maximum width of the slit ${\delta }_{\max }$ at which the fringes on the screen will be still observed sufficiently sharp.

The distances from a Fresnel biprism to a narrow slit and a screen are equal to $a=25 \mathrm{cm}$ and $b=100 \mathrm{cm}$, respectively. The refracting angle of the glass biprism is equal to $\mathrm{\theta }={20}^{\text{'}}$. Find the wavelength of light, if the width of the fringe on the screen is $\mathrm{\Delta }x=0.55 \mathrm{mm}$. For glass refractive index, $n=1.5$

A plane light wave with wavelength $\lambda =0.70 \mathrm{\mu m}$ falls normally on the base of a biprism, made of glass $(n=1.520)$ with refracting angle $\mathrm{\theta }=5.0°$. Behind the biprism there is a plane-parallel plate, with the space between them filled up with benzene $\left(n\text{'}=1.500\right)$. Find the width of a fringe on the screen $Sc$ placed behind this system.

A plane monochromatic light wave falls normally on a diaphragm with two narrow slits, separated by a distance $d=2.5 \mathrm{mm}$. A fringe pattern is formed on a screen placed at a distance $l=100 \mathrm{cm}$ behind the diaphragm. By what distance and in which direction will these fringes be displaced, when one of the slits is covered by a glass plate of thickness $h=10 \mathrm{\mu m}$? For glass $n=\frac{3}{2}$

Figure illustrates an interferometer used in measurement of refractive indices of transparent substances. Here $S$ is a narrow slit illuminated by a monochromatic light with wavelength $\lambda =589 \mathrm{nm}$, $1$ and $2$ are identical tubes with air of length $l=10.0 \mathrm{cm}$ each, $D$ is a diaphragm with two slits. After the air in tube $1$ was replaced with ammonia gas, the interference pattern on the screen $Sc$ was displaced upward by $N=17$ fringes. The refractive index of air is equal to $n=1.000277$. Determine the refractive index of ammonia gas.