Question 1:
easyWhen light propagates through a material medium of relative permittivity $\epsilon_r$ and relative permeability $\mu_r$, the velocity of light, v is given by : (c – velocity of light in vacuum)
(2022)
The velocity of light in a medium is $v = \frac{1}{\sqrt{\mu \epsilon}} = \frac{1}{\sqrt{\mu_0 \mu_r \epsilon_0 \epsilon_r}}$.
Since the velocity of light in vacuum is $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$, substituting this gives $v = \frac{c}{\sqrt{\epsilon_r \mu_r}}$.
Question 2:
easyFor a plane electromagnetic wave propagating in x-direction, which one of the following combination gives the correct possible directions for electric field (E) and magnetic field (B) respectively?
(2021)
The direction of propagation of an EM wave is given by $\vec{E} \times \vec{B}$. It is propagating in $+x$ direction (i.e., $\hat{i}$).
Checking option a: $(-\hat{j} + \hat{k}) \times (-\hat{j} - \hat{k}) = \hat{j} \times \hat{k} - \hat{k} \times \hat{j} = \hat{i} - (-\hat{i}) = 2\hat{i}$, which is parallel to $+x$.
Therefore, option a provides a valid combination.
Question 3:
easyLight with an average flux of $20 W/cm^2$ falls on non-reflecting surface at normal incidence having surface area $20 cm^2$. The energy received by the surface during time span of 1 minute is:
(2020)
Total power received by the surface is $P = Flux \times Area = 20 W/cm^2 \times 20 cm^2 = 400 W$.
The energy received in 1 minute ($60 s$) is $E = P \times t = 400 W \times 60 s$.
$E = 24000 J = 24 \times 10^3 J$.
Question 4:
easyThe ratio of contributions made by the electric field and magnetic field components to the intensity of an electromagnetic wave is : (c = speed of electromagnetic waves)
(2020)
In an electromagnetic wave, the energy is equally divided between the electric and magnetic fields.
The average energy density of the electric field equals that of the magnetic field ($u_E = u_B$).
Therefore, their ratio of contributions to the intensity is $1 : 1$.
Question 5:
easyThe magnetic field in an electromagnetic wave is given by, $B_y = 2 \times 10^{-7} \sin(\pi \times 10^3 x + 3\pi \times 10^{11} t) T$. Calculate the wavelength.
(2020-Covid)
Comparing the given equation with the standard wave equation $B = B_0 \sin(kx + \omega t)$, we get the wave number $k = \pi \times 10^3 m^{-1}$.
The wavelength is related to the wave number by $\lambda = \frac{2\pi}{k}$.
Substituting $k$, we get $\lambda = \frac{2\pi}{\pi \times 10^3} = 2 \times 10^{-3} m$.
Question 6:
easyAn em wave is propagating in a medium with a velocity $\vec{v} = v\hat{i}$. The instantaneous oscillating electric field of this em wave is along +y axis. Then the direction of oscillating magnetic field of the em wave will be along:
(2018)
The direction of propagation of an EM wave is given by the cross product $\vec{E} \times \vec{B}$.
Here, velocity is along $+x$ ($\hat{i}$) and electric field is along $+y$ ($\hat{j}$). We need $\hat{j} \times \vec{B} = \hat{i}$.
Since $\hat{j} \times \hat{k} = \hat{i}$, the magnetic field must be along the $+z$ direction ($\hat{k}$).
Question 7:
easyIn an electromagnetic wave in free space the root mean square value of the electric field is $E_{rms} = 6 V/m$. The peak value of the magnetic field is:
(2017-Delhi)
The peak value of the electric field is $E_0 = \sqrt{2} E_{rms} = 6\sqrt{2} V/m$.
The peak value of the magnetic field is $B_0 = \frac{E_0}{c} = \frac{6\sqrt{2}}{3 \times 10^8}$.
$B_0 = 2\sqrt{2} \times 10^{-8} \approx 2.828 \times 10^{-8} T \approx 2.83 \times 10^{-8} T$.
Question 8:
easyOut of the following options which one can be used to produce a propagating electromagnetic wave?
(2016 – I)
A stationary charge produces only a static electric field, while a charge moving at constant velocity produces a steady magnetic field along with it.
Only an accelerating (or oscillating) charge produces continuously changing electric and magnetic fields that sustain each other.
Therefore, an accelerating charge produces a propagating electromagnetic wave.
Question 9:
easyRadiation of energy ‘E’ falls normally on a perfectly reflecting surface. The momentum transferred to the surface is (C = velocity of light):
(2015)
The momentum of the incident radiation is $p = \frac{E}{C}$.
Since the surface is perfectly reflecting, the radiation reflects back with momentum $-p$.
The momentum transferred to the surface is the change in momentum: $\Delta p = p - (-p) = 2p = \frac{2E}{C}$.
Question 10:
easyLight with an energy flux of $25 \times 10^4 W/m^2$ falls on a perfectly reflecting surface at normal incidence. If the surface area is $15 cm^2$, the average force exerted on the surface is:
(2014)
Total power incident is $P = Flux \times Area = (25 \times 10^4) \times (15 \times 10^{-4}) = 375 W$.
For a perfectly reflecting surface, the radiation pressure is $P_{rad} = \frac{2I}{c}$, so the force is $F = \frac{2 P}{c}$.
$F = \frac{2 \times 375}{3 \times 10^8} = \frac{750}{3 \times 10^8} = 2.50 \times 10^{-6} N$.