Properties of Electromagnetic Waves Questions in English

(B) Infrared $(IR)$ radiations are electromagnetic waves with wavelengths longer than visible light.
They are primarily produced by the thermal agitation of atoms and molecules in a substance,which is why they are often called heat rays.
Option $A$ is true because $IR$ rays have longer wavelengths and can penetrate through haze and fog,making them useful for long-distance photography.
Option $B$ is false because $IR$ radiations arise due to the vibration and rotation of molecules,not due to inner electron transitions (which typically produce $X$-rays).
Option $C$ is true as a bolometer is a device used to detect and measure the intensity of incident electromagnetic radiation.
Option $D$ is true as the Sun emits a significant portion of its energy in the infrared spectrum.

(B) The correct answer is $B$. Electromagnetic $(EM)$ waves are transverse waves that consist of oscillating electric and magnetic fields. Unlike mechanical waves (such as sound waves),$EM$ waves do not require a material medium for their propagation and can travel through a vacuum.

(B) Polarization is a property exclusively associated with transverse waves.
$I.$ Water waves are surface waves which are a combination of transverse and longitudinal motions,but they are not strictly transverse in the sense required for polarization.
$II.$ Electromagnetic waves (created by an oscillating electric field) are transverse waves and can be polarized.
$III.$ Sound waves are longitudinal waves,whether in air or under water,and longitudinal waves cannot be polarized.
Therefore,only the wave mentioned in $II$ can be polarized.
The correct option is $B$.

(B) The velocity factor $(v.f.)$ of a transmission line is defined as the ratio of the velocity of the signal in the line to the velocity of light in a vacuum.
It is given by the formula: $v.f. = \frac{1}{\sqrt{k}}$,where $k$ is the dielectric constant of the medium.
Given that $k = 2.6$,we substitute this value into the formula:
$v.f. = \frac{1}{\sqrt{2.6}}$
$v.f. \approx \frac{1}{1.6124}$
$v.f. \approx 0.62$
Therefore,the value of $x$ is $0.62$.

(D) Short-wave radio broadcasting involves frequencies in the range of $3 \ MHz$ to $30 \ MHz$.
These waves are reflected back to the Earth by the ionosphere,which allows them to travel over long distances beyond the horizon.
This mode of propagation is known as sky wave propagation.
Therefore,the correct option is $D$.

(B) The energy flux $\varphi$ is defined as the power per unit area.
Given:
Pulse power $P = 10^{12} \ W$
Area $A = 10^{-4} \ cm^2$
Formula:
$\varphi = \frac{P}{A}$
Calculation:
$\varphi = \frac{10^{12}}{10^{-4}} = 10^{12 - (-4)} = 10^{16} \ W/cm^2$
Therefore,the correct option is $B$.

(B) The refractive index $n$ of the ionosphere is given by $n = \sqrt{1 - \frac{81N}{f^2}}$,where $N$ is the electron density in $m^{-3}$ and $f$ is the frequency in $Hz$.
Given $N = 400 \, \text{electrons}/cm^3 = 400 \times 10^6 \, \text{electrons}/m^3$ and $f = 55 \times 10^6 \, Hz$.
Calculating the term $\frac{81N}{f^2} = \frac{81 \times 400 \times 10^6}{(55 \times 10^6)^2} = \frac{32400 \times 10^6}{3025 \times 10^{12}} \approx 1.07 \times 10^{-8}$.
Since this value is extremely small,$n = \sqrt{1 - 1.07 \times 10^{-8}} \approx 1$.
According to Snell's Law,$n_1 \sin i = n_2 \sin r$. Assuming the wave travels from vacuum $(n_1 \approx 1)$ into the $D$-region $(n_2 \approx 1)$,
$1 \cdot \sin(45^\circ) = 1 \cdot \sin r$.
Therefore,$\sin r = \sin(45^\circ)$,which implies $r = 45^\circ$.

(A) The speed of light in free space is given by $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$.
The speed of light in a medium is given by $v = \frac{1}{\sqrt{\mu \varepsilon}}$.
The refractive index $n$ of a medium is defined as the ratio of the speed of light in vacuum to the speed of light in the medium:
$n = \frac{c}{v} = \frac{1/\sqrt{\mu_0 \varepsilon_0}}{1/\sqrt{\mu \varepsilon}} = \sqrt{\frac{\mu \varepsilon}{\mu_0 \varepsilon_0}}$.

(C) The speed of light in a vacuum is given by $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$.
The speed of light in a medium is given by $v = \frac{1}{\sqrt{\mu \varepsilon}}$,where $\mu = \mu_0 \mu_r$ and $\varepsilon = \varepsilon_0 \varepsilon_r = \varepsilon_0 K$.
Here,$\mu_r$ is the relative permeability and $K$ is the dielectric constant (relative permittivity).
The refractive index $n$ is defined as the ratio of the speed of light in a vacuum to the speed of light in the medium:
$n = \frac{c}{v} = \frac{\frac{1}{\sqrt{\mu_0 \varepsilon_0}}}{\frac{1}{\sqrt{\mu \varepsilon}}} = \sqrt{\frac{\mu \varepsilon}{\mu_0 \varepsilon_0}} = \sqrt{\frac{\mu_0 \mu_r \varepsilon_0 K}{\mu_0 \varepsilon_0}} = \sqrt{\mu_r K}$.
Therefore,the correct option is $C$.

(C) In vacuum,the speed of light is a universal constant,denoted by $c$.
It is given by the relation $c = f \lambda$,where $f$ is the frequency and $\lambda$ is the wavelength.
Since $c$ is constant $(3 \times 10^8 \ m/s)$,it does not depend on the frequency or the wavelength of the electromagnetic wave.
Therefore,the speed of light in a vacuum is independent of both frequency and wavelength.
Thus,option $C$ is the correct statement.

(D) In vacuum,the speed of light is a universal constant,denoted by $c$.
Its value is approximately $3 \times 10^8 \ m/s$.
It does not depend on the frequency,wavelength,or the velocity of the source of light.
Therefore,the correct option is $D$.

(B) The relationship between the speed of light $(c)$,frequency $(\nu)$,and wavelength $(\lambda)$ is given by the formula: $c = \nu \lambda$.
Rearranging for wavelength,we get: $\lambda = \frac{c}{\nu}$.
Given that the speed of light $c = 3 \times 10^8\;m/s$ and the frequency $\nu = 100\;Hz$,we substitute these values into the equation:
$\lambda = \frac{3 \times 10^8}{100} = 3 \times 10^6\;m$.
Therefore,the correct option is $B$.

(A) Light is an electromagnetic wave in nature.
Electromagnetic waves do not require any material medium for their propagation.
Therefore,the statement that light requires a material medium for propagation is incorrect.
Thus,option $A$ is the correct answer.

(C) According to the postulates of the Special Theory of Relativity,the speed of light in a vacuum is an absolute constant,denoted by $c$.
This speed is independent of the motion of the source or the motion of the observer.
Since all three observers $A, B$ and $C$ are measuring the speed of light in a vacuum,they will all measure the same value,regardless of their relative velocities.
Therefore,${v_A} = {v_B} = {v_C} = c$.

(A) The plane of polarisation is defined as the plane that contains the direction of propagation of the electromagnetic wave and the electric field vector $\overrightarrow{E}$.
Since the plane of polarisation contains the direction of propagation,the angle between the direction of propagation and the plane of polarisation is $0^o$.

(D) The speed of an electromagnetic wave in vacuum is related to the permeability of free space $(\mu_0)$ and the permittivity of free space $(\varepsilon_0)$ by the Maxwell's relation:
The speed of light $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$.
Given values are $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$ and $\varepsilon_0 \approx 8.85 \times 10^{-12} \text{ C}^2/(\text{N m}^2)$.
Substituting these values,we get $c \approx 3 \times 10^8 \text{ m/s}$.

(C) The speed of electromagnetic $(EM)$ waves in a vacuum is given by the formula $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$.
Here,$\mu_0$ is the permeability of free space and $\epsilon_0$ is the permittivity of free space.
Since both $\mu_0$ and $\epsilon_0$ are universal constants,the speed of $EM$ waves in a vacuum is a constant value $(c \approx 3 \times 10^8 \ m/s)$ for all frequencies and wavelengths,regardless of the source of radiation.
Therefore,the speed is the same for all electromagnetic waves.

(B) The velocity of electromagnetic $(EM)$ waves in a vacuum is given by the formula $v = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$.
Substituting the values of permeability of free space $(\mu_0 = 4\pi \times 10^{-7} \ T \cdot m/A)$ and permittivity of free space $(\epsilon_0 = 8.854 \times 10^{-12} \ C^2/N \cdot m^2)$,we get $v \approx 3 \times 10^8 \ m/s$.
This value is exactly equal to the speed of light in a vacuum.
Therefore,electromagnetic waves travel with a velocity equal to the velocity of light.

(B) The greenhouse effect is primarily caused by infrared radiations.
Solar energy reaches the Earth's surface,which then re-emits this energy as infrared radiation.
Greenhouse gases in the atmosphere absorb these infrared rays and reflect them back towards the Earth's surface,thereby trapping heat and keeping the planet warm.

(A) The transverse nature of electromagnetic waves is confirmed by the phenomenon of polarization.
Polarization is a property that is exclusive to transverse waves,as it involves the restriction of the oscillation of the electric field vector to a specific plane perpendicular to the direction of wave propagation.
Longitudinal waves,such as sound waves,cannot be polarized because their oscillations occur along the direction of propagation.
Therefore,the correct option is $A$.

(C) The electromagnetic wave propagates in a direction perpendicular to both the electric field vector $\overrightarrow E$ and the magnetic field vector $\overrightarrow B$.
According to the properties of electromagnetic waves,the direction of wave propagation is given by the direction of the Poynting vector $\vec S$,which is proportional to $\overrightarrow E \times \overrightarrow B$.
Therefore,the direction of propagation is along $\overrightarrow E \times \overrightarrow B$.

(A) Ozone $(O_3)$ is present in the stratosphere region of the atmosphere.
It protects life on Earth by absorbing harmful ultraviolet $(UV)$ radiation emitted by the sun.
Exposure to high levels of $UV$ radiation causes various health problems in humans,such as skin cancer and cataracts,and damages plants and other animals,leading to ecosystem disruption.
Plants cannot grow effectively under intense ultraviolet radiation.
Therefore,the primary biological importance of the ozone layer is the absorption of these harmful rays.
Hence,option $A$ is correct.

(A) In vacuum, all electromagnetic $(EM)$ waves travel with the same speed, which is the speed of light, $c \approx 3 \times 10^8 \ m/s$.
Since the frequency $(f)$ and wavelength $(\lambda)$ are related by the equation $c = f \lambda$, and different types of electromagnetic waves have different frequencies, they must have different wavelengths.
Therefore, radio waves and visible light have the same velocity but different wavelengths.

(C) In an electromagnetic wave,energy is carried by both the oscillating electric field and the oscillating magnetic field.
The energy density associated with the electric field is given by $u_E = \frac{1}{2} \varepsilon_0 E^2$.
The energy density associated with the magnetic field is given by $u_B = \frac{1}{2} \frac{B^2}{\mu_0}$.
The total energy density $u$ is the sum of these two: $u = u_E + u_B = \frac{1}{2} \varepsilon_0 E^2 + \frac{1}{2} \frac{B^2}{\mu_0}$.
Therefore,the energy stored in electromagnetic oscillations is in the form of both electrical and magnetic energy.

(A) The relationship between the speed of light $(c)$,frequency $(\nu)$,and wavelength $(\lambda)$ is given by the formula: $c = \nu \lambda$.
Rearranging for wavelength,we get: $\lambda = \frac{c}{\nu}$.
Given: $c = 3 \times 10^8 \ m/s$ and $\nu = 8.2 \times 10^6 \ Hz$.
Substituting the values: $\lambda = \frac{3 \times 10^8}{8.2 \times 10^6} = \frac{300}{8.2} \approx 36.58 \ m$.
Rounding to the nearest provided option,we get $36.6 \ m$.

(B) The relationship between the amplitude of the electric field $(E_0)$ and the amplitude of the magnetic field $(B_0)$ in an electromagnetic wave is given by the equation: $c = \frac{E_0}{B_0}$, where $c$ is the speed of light in vacuum $(c \approx 3 \times 10^8 \text{ m/s})$.
Rearranging the formula to solve for $B_0$, we get: $B_0 = \frac{E_0}{c}$.
Given $E_0 = 18 \text{ V/m}$ and $c = 3 \times 10^8 \text{ m/s}$, we substitute these values:
$B_0 = \frac{18}{3 \times 10^8} = 6 \times 10^{-8} \text{ T}$.
Thus, the magnitude of the oscillating magnetic field is $6 \times 10^{-8} \text{ T}$.

(C) In an electromagnetic wave,the oscillating electric field vector $\vec{E}$ and the magnetic field vector $\vec{B}$ are always mutually perpendicular to each other and also perpendicular to the direction of wave propagation.
Furthermore,these fields are in phase,which means they reach their maximum and minimum values at the same time and at the same spatial location.
Therefore,the correct option is $C$.

(A) An electromagnetic wave travels in the direction of the Poynting vector $\vec{S} = \frac{1}{\mu_0} (\vec{E} \times \vec{B})$.
Given that the wave travels along the $z$-axis,the direction of propagation is $\hat{k}$.
We know that $\vec{E} \times \vec{B}$ must be in the direction of $\hat{k}$.
For option $A$,$\vec{E} = E_x \hat{i}$ and $\vec{B} = B_y \hat{j}$.
Calculating the cross product: $\hat{i} \times \hat{j} = \hat{k}$.
Thus,$E_x \hat{i} \times B_y \hat{j} = (E_x B_y) \hat{k}$,which is along the $z$-axis.
Therefore,the pair $(E_x, B_y)$ generates an electromagnetic wave traveling along the $z$-axis.

(B) Electromagnetic $(EM)$ waves are oscillations of electric and magnetic fields. They carry energy,momentum,and information through space. However,since electromagnetic waves consist of oscillating fields and not particles with net charge,they do not transport charge. Therefore,the correct option is $(B)$.

(B) Electromagnetic $(EM)$ waves consist of oscillating electric and magnetic fields that propagate through space.
These waves carry both energy and momentum.
When an $EM$ wave is incident on a material surface,it transfers energy to the surface and exerts radiation pressure due to the momentum it carries.
Therefore,for an incident $EM$ wave,both the delivered momentum $p$ and the delivered energy $E$ are non-zero.
Thus,$p \neq 0$ and $E \neq 0$.

(C) The angular wave number is given by $k = \frac{2\pi}{\lambda}$,where $\lambda$ is the wavelength.
The angular frequency is given by $\omega = 2\pi\nu$,where $\nu$ is the frequency.
The ratio $\frac{k}{\omega} = \frac{2\pi/\lambda}{2\pi\nu} = \frac{1}{\lambda\nu}$.
Since the speed of light in vacuum is $c = \lambda\nu$,we have $\frac{k}{\omega} = \frac{1}{c}$.
Since $c$ is a constant,the ratio $\frac{k}{\omega}$ is independent of the wavelength $\lambda$.

(A) For an electromagnetic wave in a vacuum,the relationship between the amplitudes of the electric field $(E_0)$ and the magnetic field $(B_0)$ is given by $E_0 = c B_0$,where $c$ is the speed of light in a vacuum.
We know that the speed of light $c$ is related to the angular frequency $\omega$ and the wave number $k$ by the equation $c = \frac{\omega}{k}$.
Substituting this into the first relation,we get $E_0 = \left( \frac{\omega}{k} \right) B_0$.
Rearranging this equation,we obtain $E_0 k = B_0 \omega$.
Therefore,the correct option is $A$.

(A) The intensity $I$ of an electromagnetic wave is given by $I = \frac{1}{2} \epsilon_0 c E_0^2$,where $E_0$ is the amplitude of the electric field.
Rearranging for $E_0$: $E_0 = \sqrt{\frac{2I}{\epsilon_0 c}}$.
Given $I = 5 \times 10^{-16} \, W/m^2$,$\epsilon_0 = 8.854 \times 10^{-12} \, F/m$,and $c = 3 \times 10^8 \, m/s$:
$E_0 = \sqrt{\frac{2 \times 5 \times 10^{-16}}{8.854 \times 10^{-12} \times 3 \times 10^8}} = \sqrt{\frac{10 \times 10^{-16}}{26.562 \times 10^{-4}}} = \sqrt{0.376 \times 10^{-12}} \approx 0.613 \times 10^{-6} \, V/m$.
The maximum potential difference $V_0$ across an antenna of length $L = 2 \, m$ is $V_0 = E_0 \times L$.
$V_0 = 0.613 \times 10^{-6} \times 2 = 1.226 \times 10^{-6} \, V \approx 1.23 \, \mu V$.

(C) The refractive index $n$ of a medium is given by $n = \sqrt{\frac{\mu\varepsilon}{\mu_0\varepsilon_0}}$.
Assuming the medium is non-magnetic,$\mu = \mu_0$,so $n = \sqrt{\frac{\varepsilon}{\varepsilon_0}}$.
Given the relative permittivity (dielectric constant) $K = \frac{\varepsilon}{\varepsilon_0} = 4.0$,we have $n = \sqrt{4.0} = 2$.
The frequency $\nu$ of an electromagnetic wave depends only on the source and remains unchanged when passing through different media.
The speed of the wave in the medium is $v = \frac{c}{n} = \frac{c}{2}$.
Since $v = \nu\lambda$,the new wavelength $\lambda' = \frac{v}{\nu} = \frac{c/2}{\nu} = \frac{\lambda}{2}$.
Thus,the wavelength is halved and the frequency remains unchanged.

(D) The speed of an electromagnetic wave in a medium is given by the formula $v = \frac{c}{\sqrt{\mu_r \varepsilon_r}}$,where $c$ is the speed of light in vacuum $(3 \times 10^8 \ m/s)$,$\mu_r$ is the relative permeability,and $\varepsilon_r$ is the relative permittivity.
Given: $\mu_r = 1.3$ and $\varepsilon_r = 2.14$.
Substituting the values:
$v = \frac{3 \times 10^8}{\sqrt{1.3 \times 2.14}}$
$v = \frac{3 \times 10^8}{\sqrt{2.782}}$
$v \approx \frac{3 \times 10^8}{1.668}$
$v \approx 1.8 \times 10^8 \ m/s$.
Therefore,the correct option is $D$.

(C) The momentum of a photon is given by $p = E/c$.
The intensity $I$ is defined as the energy incident per unit area per unit time,$I = E/(At)$.
For a non-reflecting (perfectly absorbing) surface,the momentum transferred per unit area per unit time is equal to the radiation pressure $P$.
Since the photon is absorbed,the change in momentum per unit area per unit time is $\Delta p / (At) = (E/c) / (At) = I/c$.
Therefore,the radiation pressure exerted on a non-reflecting surface is $P = I/c$.

(C) The direction of propagation of an electromagnetic wave is given by the direction of the vector $\vec E \times \vec B$.
Given that the wave travels along the $y$-direction (unit vector $\hat j$) and the electric field $\vec E$ is along the $x$-axis (unit vector $\hat i$).
Let the direction of the magnetic field $\vec B$ be $\hat n$.
Then,$\hat i \times \hat n = \hat j$.
We know that $\hat i \times \hat k = -\hat j$ and $\hat i \times (-\hat k) = \hat j$.
Therefore,the direction of the $\vec B$ vector must be along the $+z$-axis. Note: The original option $(d)$ was $-z$-axis,but mathematically $\hat i \times \hat k = \hat j$ is incorrect; $\hat i \times \hat k = -\hat j$. Thus,for $\hat i \times \hat n = \hat j$,$\hat n$ must be $-\hat k$. Let us re-evaluate: $\hat i \times \hat k = \hat j$ is false,$\hat i \times \hat j = \hat k$,$\hat j \times \hat k = \hat i$,$\hat k \times \hat i = \hat j$. So,$\hat k \times \hat i = \hat j$. If $\vec E$ is along $x$ $(\hat i)$ and propagation is along $y$ $(\hat j)$,then $\vec E \times \vec B = \hat i \times \vec B = \hat j$. Since $\hat k \times \hat i = \hat j$,$\vec B$ must be along the $+z$-axis.

(C) The speed of electromagnetic waves in a vacuum is given by $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$.
In a medium with absolute permeability $\mu$ and absolute permittivity $\varepsilon$,the speed is $v = \frac{1}{\sqrt{\mu \varepsilon}}$.
We know that $\mu = \mu_0 \mu_r$ and $\varepsilon = \varepsilon_0 K$,where $\mu_r$ is the relative permeability and $K$ is the dielectric constant (relative permittivity).
Substituting these into the expression for $v$,we get $v = \frac{1}{\sqrt{(\mu_0 \mu_r)(\varepsilon_0 K)}} = \frac{1}{\sqrt{\mu_0 \varepsilon_0} \sqrt{\mu_r K}}$.
Since $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$,we can write $v = \frac{c}{\sqrt{\mu_r K}}$.

(B) The average energy density of the electric field $(u_e)$ in an electromagnetic wave is given by the formula:
$u_e = \frac{1}{4} \varepsilon_0 E_0^2$
Where $\varepsilon_0$ is the permittivity of free space $(8.85 \times 10^{-12} \ F/m)$ and $E_0$ is the amplitude of the electric field.
Given $E_0 = 1 \ V/m$:
$u_e = \frac{1}{4} \times 8.85 \times 10^{-12} \times (1)^2$
$u_e = 2.2125 \times 10^{-12} \ J/m^3$
Rounding to the nearest option,we get $2.2 \times 10^{-12} \ J/m^3$.

(A) The power radiated as electromagnetic waves is $P = 3\% \text{ of } 100\,W = 3\,W$.
The intensity $S_{av}$ at a distance $R = 10\,m$ is given by $S_{av} = \frac{P}{4\pi R^2}$.
Also,the intensity of an electromagnetic wave is related to the electric field amplitude $E_0$ by $S_{av} = \frac{1}{2} \varepsilon_0 c E_0^2$.
Equating the two expressions: $\frac{1}{2} \varepsilon_0 c E_0^2 = \frac{P}{4\pi R^2}$.
Solving for $E_0$: $E_0 = \sqrt{\frac{P}{2\pi R^2 \varepsilon_0 c}}$.
Substituting the values: $E_0 = \sqrt{\frac{3}{2 \times 3.14 \times (10)^2 \times 8.85 \times 10^{-12} \times 3 \times 10^8}}$.
$E_0 = \sqrt{\frac{3}{1665.72 \times 10^{-4}}} \approx \sqrt{1.8} \approx 1.34\,V/m$.

(D) The intensity $I$ of an electromagnetic wave at a distance $R$ from a point source is given by $I = \frac{P}{4\pi R^2}$.
Also,the intensity in terms of the maximum electric field $E_0$ is $I = \frac{1}{2} \varepsilon_0 E_0^2 c$.
Equating the two expressions: $\frac{P}{4\pi R^2} = \frac{1}{2} \varepsilon_0 E_0^2 c$.
Solving for $E_0$: $E_0 = \sqrt{\frac{P}{2\pi R^2 \varepsilon_0 c}}$.
Substituting the given values: $P = 800 \, W$,$R = 4.0 \, m$,$\varepsilon_0 = 8.85 \times 10^{-12} \, F/m$,and $c = 3 \times 10^8 \, m/s$.
$E_0 = \sqrt{\frac{800}{2 \times 3.14 \times (4)^2 \times 8.85 \times 10^{-12} \times 3 \times 10^8}}$.
$E_0 = \sqrt{\frac{800}{1331.328 \times 10^{-4}}} = \sqrt{6008.9} \approx 54.77 \, V/m$.

(C) The wave impedance $Z$ of a medium is given by the formula $Z = \sqrt{\frac{\mu}{\varepsilon}} = \sqrt{\frac{\mu_r \mu_0}{\varepsilon_r \varepsilon_0}}$.
Given relative permittivity $\varepsilon_r = 2$ and relative permeability $\mu_r = 50$.
The impedance of free space is $Z_0 = \sqrt{\frac{\mu_0}{\varepsilon_0}} \approx 376.6 \, \Omega$.
Substituting the values:
$Z = \sqrt{\frac{50}{2}} \times 376.6$
$Z = \sqrt{25} \times 376.6$
$Z = 5 \times 376.6 = 1883 \, \Omega$.

(D) The intensity of the wave is $I = 6 \ W/m^2$ and the area of the mirror is $A = 40 \ cm^2 = 40 \times 10^{-4} \ m^2$.
For a perfectly reflecting surface (mirror) held perpendicular to the wave,the radiation pressure exerts a force,and the momentum transferred per second is equal to the force exerted.
The momentum transferred per second is given by $p = \frac{2U}{c}$,where $U$ is the energy incident per second $(U = I \times A)$ and $c$ is the speed of light $(3 \times 10^8 \ m/s)$.
Substituting the values:
$p = \frac{2 \times I \times A}{c} = \frac{2 \times 6 \times 40 \times 10^{-4}}{3 \times 10^8}$
$p = \frac{480 \times 10^{-4}}{3 \times 10^8} = 160 \times 10^{-12} = 1.6 \times 10^{-10} \ kg \cdot m/s$.

(B) The critical frequency $f_c$ of the ionosphere is given by the formula $f_c = 9 \sqrt{N_{\max}}$,where $N_{\max}$ is the maximum electron density.
Let $N_1 = 10^{10} \ m^{-3}$ be the density in the morning and $N_2 = 2 \times 10^{10} \ m^{-3}$ be the density at noon.
The critical frequency in the morning is $f_{c1} = 9 \sqrt{N_1}$.
The critical frequency at noon is $f_{c2} = 9 \sqrt{N_2}$.
The ratio of the critical frequency at noon to the critical frequency in the morning is $\frac{f_{c2}}{f_{c1}} = \frac{9 \sqrt{N_2}}{9 \sqrt{N_1}} = \sqrt{\frac{N_2}{N_1}}$.
Substituting the values: $\frac{f_{c2}}{f_{c1}} = \sqrt{\frac{2 \times 10^{10}}{10^{10}}} = \sqrt{2} \approx 1.414$.

(C) The relationship between velocity $(c)$,frequency $(f)$,and wavelength $(\lambda)$ is given by the formula: $c = f \lambda$.
Rearranging to solve for frequency: $f = \frac{c}{\lambda}$.
Given: $c = 3 \times 10^{8} \ m/s$ and $\lambda = 150 \ m$.
Substituting the values: $f = \frac{3 \times 10^{8}}{150} = \frac{300 \times 10^{6}}{150} = 2 \times 10^{6} \ Hz$.
Since $10^{6} \ Hz = 1 \ MHz$,the frequency is $2 \ MHz$.

(B) The refractive index $n$ of the ionosphere is given by $n = \sqrt{1 - \frac{80.5 N}{f^2}}$, where $N$ is the electron density in $\text{m}^{-3}$ and $f$ is the frequency in $\text{Hz}$.
Given: $f = 55 \times 10^6 \text{ Hz}$, $N = 400 \text{ electrons/cm}^3 = 400 \times 10^6 \text{ electrons/m}^3$.
Calculating the term inside the square root: $\frac{80.5 \times 400 \times 10^6}{(55 \times 10^6)^2} = \frac{32200 \times 10^6}{3025 \times 10^{12}} \approx 1.06 \times 10^{-8}$.
Since this value is extremely small compared to $1$, $n \approx \sqrt{1 - 0} = 1$.
Using Snell's Law: $n_1 \sin i = n_2 \sin r$. Assuming the wave travels from vacuum $(n_1 \approx 1)$ into the ionosphere $(n_2 \approx 1)$:
$1 \cdot \sin(45^\circ) = 1 \cdot \sin r$.
Therefore, $r = 45^\circ$.