Electromagnetic waves with wavelength
(i) $\lambda_1$ is used in satellite communication.
(ii) $\lambda_2$ is used to kill germs in water purifies.
(iii) $\lambda_3$ is used to detect leakage of oil in underground pipelines.
(iv) $\lambda_4$ is used to improve visibility in runways during fog and mist conditions.
(a) Identify and name the part of electromagnetic spectrum to which these radiations belong.
(b) Arrange these wavelengths in ascending order of their magnitude.
(c) Write one more application of each.
(a) (i) Microwave is used in satellite communications.
So, $\lambda_1$ is the wavelength of microwave.
(ii) Ultraviolet rays are used to kill germs in water purifier. So, $\lambda_2$ is the wavelength of UV rays.
(iii) $X$-rays are used to detect leakage of oil in underground pipelines. So, $\lambda_3$ is the wavelength of $X$-rays.
(iv) Infrared is used to improve visibility on runways during fog and mist conditions. So, it is the wavelength of infrared waves.
(b) Wavelength of X-rays < wavelength of UV $<$ wavelength of infrared $<$ wavelength of microwave.
$$\Rightarrow \quad \lambda_3<\lambda_2<\lambda_4<\lambda_1 $$
(c) (i) Microwave is used in radar.
(ii) UV is used in LASIK eye surgery.
(iii) X-ray is used to detect a fracture in bones.
(iv) Infrared is used in optical communication.
Show that average value of radiant flux density $S$ over a single period $T$ is given by $S=\frac{1}{2 c \propto_0} E_0^2$.
Radiant flux density $S=\frac{1}{\propto_0}(\mathbf{E} \times \mathbf{B})=C^2 \varepsilon_0(\mathbf{E} \times \mathbf{B})$
$$\left[\because c=\frac{1}{\sqrt{\propto_0 \varepsilon_0}}\right]$$
Suppose electromagnetic waves be propagating along $x$-axis. The electric field vector of electromagnetic wave be along $y$-axis and magnetic field vector be along $z$-axis. Therefore,
$$\begin{aligned} \mathbf{E}_0 & =\mathbf{E}_0 \cos (k x-\omega t) \\ \text{and}\quad\mathbf{B} & =\mathbf{B}_0 \cos (k x-\omega t) \\ \mathbf{E} \times \mathbf{B} & =\left(\mathbf{E}_0 \times \mathbf{B}_0\right) \cos ^2(k x-\omega t) \\ \mathbf{S} & =\mathbf{c}^2 \varepsilon_0(\mathbf{E} \times \mathbf{B}) \\ & =c^2 \varepsilon_0\left(\mathbf{E}_0 \times \mathbf{B}_0\right) \cos ^2(k x-\omega t) \end{aligned}$$
Average value of the magnitude of radiant flux density over complete cycle is
$$\begin{array}{rlrl} S_{\mathrm{av}} & =c^2 \varepsilon_{\mathrm{o}}\left|\mathbf{E}_0 \times \mathbf{B}_0\right| \frac{1}{T} \int_0^T \cos ^2(k x-\omega t) d t \\ & =c^2 \varepsilon_0 E_0 B_0 \times \frac{1}{T} \times \frac{T}{2} \quad\left[\because \int_0^T \cos ^2(k x-\omega t) d t=\frac{T}{2}\right] \\ \Rightarrow \quad S_{\mathrm{av}} & =\frac{c^2}{2} \varepsilon_0 E_0\left(\frac{E_0}{c}\right) \quad\left[\text { As, } C=\frac{E_0}{B_0}\right]\\ & =\frac{c}{2} \varepsilon_0 E_0^2=\frac{c}{2} \times \frac{1}{c^2 \alpha_0} E_0^2 \quad {\left[C=\frac{1}{\sqrt{\alpha_0 \varepsilon_0}} \text { or } \varepsilon_0=\frac{1}{c^2 \alpha_0}\right]}\\ \Rightarrow \quad S_{\mathrm{av}} & =\frac{E_0^2}{2 \alpha_0 c} \end{array}$$
Hence proved.
You are given a $2 \propto F$ parallel plate capacitor. How would you establish an instantaneous displacement current of 1 mA in the space between its plates?
Given, capacitance of capacitor $C=2 \propto F$,
Displacement current $I_d=1 \mathrm{~mA}$
Charge $$q=C V$$
$$\begin{array}{rlr} I_d d t & =C d V & {[\because q=i t]} \\ \text{or} \quad I_d & =C \frac{d V}{d t} \\ 1 \times 10^{-3} & =2 \times 10^{-6} \times \frac{d V}{d t} \\ \text{or}\quad\frac{d V}{d t} & =\frac{1}{2} \times 10^{+3}=500 \mathrm{~V} / \mathrm{s} & \end{array}$$
So, by applying a varying potential difference of 500 V/s, we would produce a displacement current of desired value.
Show that the radiation pressure exerted by an EM wave of intensity $I$ on a surface kept in vacuum is $\frac{I}{C}$.
$$\text { Pressure }=\frac{\text { Force }}{\text { Area }}=\frac{F}{A}$$
Force is the rate of change of momentum i.e.,
$$\begin{aligned} \text{i.e.,}\quad F & =\frac{d p}{d t} \\ \text{Energy in time dt,}\quad U & =p \cdot C \text { or } p=\frac{U}{C} \\ \therefore\quad\text { Pressure } & =\frac{1}{A} \cdot \frac{U}{C \cdot d t} \\ \text { Pressure } & =\frac{I}{C} \quad\left[\because I=\text { Intensity }=\frac{U}{A \cdot d t}\right] \end{aligned}$$
What happens to the intensity of light from a bulb if the distance from the bulb is doubled? As a laser beam travels across the length of room, its intensity essentially remain constant. What geometrical characteristic of LASER beam is responsible for the constant intensity which is missing in the case of light from the bulb?
As the distance is doubled, the area of spherical region $\left(4 \pi r^2\right)$ will become four times, so the intensity becomes one fourth the initial value $\left(\because I \propto \frac{1}{r^2}\right)$ but in case of laser it does not spread, so its intensity remain same.Geometrical characteristic of LASER beam which is responsible for the constant intensity are as following
(i) Unidirection
(ii) Monochromatic
(iii) Coherent light
(iv) Highly collimated
These characteristic are missing in the case of light from the bulb.