Physics M.C.Q (1 Marks)

3. 5.405 × 10¹⁴ Hz

Explanation:

Human eye is sensitive to light of wavelength ➔ $\lambda$ = 5550 angstrom.

So its frequency is $\text{v}=\frac{\text{c}}{\lambda}$

$\text{v} =\frac{5550}{3 × 10^8} × 10^{-10}$

v = 5.405 × 10¹⁴ Hz

1. E_y, B_z.

4. E_z , B_y.

​​​​​​​Solution:

Key concept: The direction of propagation of electromagnetic wave is perpendicular to both electric field vector $(\vec{\text{E}})$ and $(\vec{\text{B}})$ magntic field vector B, i.e., in the direction of $\vec{\text{E}}\times\vec{\text{B}}$.

Here in the question electromagnetic wave is propagating along x-direction. So, electro and magnetic field vectors should have either y-direction of 2-direction.

2. Is dependent on nature of surface and intensity of light used

Explanation:

Radiation pressure is given by $\text{P}_\text{R}=\frac{(1+\alpha)\text{I}}{\text{C}}$

where α is the coefficient of reflection of the surface.

For completely reflecting surface $\alpha=1$

For completely absorbing surface $\alpha=0$

So, radiation pressure depends on the nature of surface on which the light is falling but independent of wavelength of light falling.

3. For all intensities.

Explanation:

For any given medium, the speed (c) of an electromagnetic wave is given by,

$\text{C}=\text{v}\lambda$

Where,

V = Frequency of the electromagnetic wave.

$\lambda=$ wavelength of the electromagnetic wave.

As the frequency and wavelength are changed, the speed of the electromagnetic wave changes. So, the speed of an electromagnetic wave is not same for all wavelengths and all frequencies in any medium. The velocity of an electromagnetic wave changes with change in medium. Also, the speed of an electromagnetic wave is same for all the intensities in any medium.

4. For an electromagnetic wave propagating in +x direction the electric field is $\overrightarrow{\text{E}}=\frac{1}{\sqrt{2}}\text{E}_\text{yz}(\text{x,t})(\hat{\text{y}}-\hat{\text{z}})$ and the magnetic field is $\overrightarrow{\text{B}}=\frac{1}{\sqrt{2}}\text{E}_\text{yz}(\text{x,t})(\hat{\text{y}}+\hat{\text{z}})$

Explanation:

Electromagnetic waves travel in the direction perpendicular to electric as well as magnetic field. Cross product of electric and magnetic field should give the direction of electromagnetic wave.

1. Electric field.
2. Magnetic field.

Explanation:

In a plane electromagnetic wave, the electric and the magnetic fields oscillate sinusoidally. For an electromagnetic wave propagating in the z-direction, the electric and magnetic fields are given by,

$\text{E}_\text{x}=\text{E}_0\sin(\text{kz}-\omega\text{t})$

$\text{B}_\text{y}=\text{B}_0\sin(\text{kz}-\omega\text{t})$

These are sinusoidal functions. Therefore, for a fixed value of z. the average value of the electric and magnetic fields are zero.

2. $\frac{2000\text{Volt}}{\text{meter}}$

Explanation:

As we know, $\text{E}\propto\text{R}^{-2}$

Therefore, $\text{E}(\text{R}=5)=\frac{500}{10^{-2}}5^{-2}=\frac{2000\text{V}}{\text{m}}$

3. $\frac{1}{\text{r}}$

Solution:

A diode antenna radiates the electromagnetic waves outwards. The amplitude of electric field vector (E₀) which transports significant energy from the source falls intensity inversely as the distance (r) from the antenna,

i.e., $\text{E}_0\propto\frac{1}{\text{r}}$.

3. $\frac{10^6\text{V}}{\text{s}}$

Explanation:

In a capacitor of capacitance C,

$\text{V}=\frac{\text{q}}{\text{C}}$

$\Rightarrow\frac{\text{dV}}{\text{dt}}=\frac{\text{i}}{\text{C}}=\frac{1\text{A}}{1\mu\text{F}}=\frac{10^6\text{V}}{\text{s}}$

3. 6.28mm

Explanation:

The general equation of an electromagnetic wave is $\text{B = A}\sin(\text{kx + }\omega\text{t})$

Comparing this equation with the given equation, A = 2 × 10^-7, k = 10³ and $\omega$ = 1.5 × 10¹²

So, $10^3=\frac{2\pi}{\lambda}$

or $\lambda=6.28\times10^{-3}\text{m}=6.28\text{mm}$

1. 3.1×10⁻⁸ along positive z-direction

Explanation:

$\text{B}=\frac{\text{E}}{\text{C}}$

$=\frac{9.3}{3\times10^8}$

$=3.1\times10^{-8}$

1. Radiation pressure is I/c if the wave is totally absorbed.

3. Radiation pressure is 2I/c if the wave is totally reflected.
4. Radiation pressure is in the range I/c < p < 2I/c for real surfaces.

Solution:

Key concept: Radiation pressure (p) is the force exerted by electromagnetic wave on unit area of the surface, i.e., rate of change of momentum per unit area of the surface.

Let us consider a surface exposed to electromagnetic radiation as shown in figure. The radiation is falling normally on the surface. Further, intensity of radiation is I and area of surface exposed to radiation is A.

E = Energy received by surface per second = I.A

N = Number of photons received by surface per second

$\text{N}=\frac{\text{E}}{\text{E}_\text{Photon}}=\frac{\text{E}\lambda}{\text{hc}}=\frac{\text{IA}\lambda}{\text{hc}}$

Now, there are three cases possible which are as follows.

CaseI:

Surface is perfectly reflecting

$\Delta\text{P}_\text{one photon}=\text{Change in momentum}=\frac{2\text{h}}{\lambda}$

$\therefore\ \text{Total force experienced F}=\text{N}\times\Delta\text{P}_\text{one photon}=\frac{2\text{IA}}{\text{c}}$

Also, pressure $\text{P}=\frac{\text{F}}{\text{A}}=\frac{2\text{I}}{\lambda}$

Case II:

Surface is perfectly absorbing

$\Delta\text{P}_\text{one photon}=\frac{\text{h}}{\lambda}$

$\Rightarrow\ \text{F}=\text{N}\times\Delta\text{P}_\text{one photon}=\frac{\text{IA}}{\text{c}}$

Also, Pressure $\text{P}=\frac{\text{F}}{\text{A}}=\frac{\text{I}}{\text{c}}$

Hence radiation pressure is in the range $\frac{\text{I}}{\text{C}}<\text{P}<\frac{2\text{I}}{\text{c}}$ for real surfaces.

Important Points:

If surface is partly reflecting

Let us consider that surface reflects 70% and absorbs 30% of the incident radiation.

$\text{F}=0.7\Big(\frac{2\text{IA}}{\text{c}}\Big)+0.3\Big(\frac{\text{IA}}{\text{c}}\Big)=\frac{1.7\text{IA}}{\text{c}}$

Remarks:

1. Radiation force/pressure supports photon theory of radiation.
2. If radiation falls abliquely, then appropriate projection of area vector is taken.

For situation as shown in figure,

$\text{F}=\frac{2\text{IA}\cos^2\theta}{\text{c}}$, for perfectly reflecting surface

$\text{F}=\frac{\text{IA}\cos\theta}{\text{c}}$, for perfectly absorbing surface

$\text{F}=\frac{1.4\text{IA}\cos^2\theta}{\text{c}}+\frac{0.3\text{IA}\cos \theta}{\text{c}}$, for partially reflecting surface.

3. $\frac{\text{k}}{\omega}$

Explanation:

$\text{k}=\frac{2\pi}{\lambda}$

$\omega=\frac{2\pi\text{c}}{\lambda},$ where c is the velocity of light 

Hence, $=\frac{\text{k}}{2\pi}=\frac{\omega}{2\pi\text{c}}$

$\Rightarrow\frac{\text{k}}{\omega}$ is independent of the wavelength.

2. $\frac{1}{\sqrt\mu\in}$

Explanation:

Velocity of light in a medium,

$\text{c}=\frac{1}{\sqrt{\mu_0\in_0\mu_\text{r}\in_\text{r}}}=\frac{1}{\mu_0\in_0}$

1. Increases.
2. Decreases.

Explanation:

Displacement current inside a capacitor,

$\text{i}_\text{d}=\in_0\frac{\text{d}\phi_\text{E}}{\text{dt}},$ where

$\phi_\text{E}$ is the electric flux inside the capacitor.

Up to the time the electric flux changes, there will be a displacement current. This is possible when the charge on a capacitor changes. Therefore, the displacement current goes through the gap between the plates of a capacitor when the charge of the capacitor or electric field increases or decreases inside the capacitor.

4. None of the above is correct

Explanation:

As we know, velocity of electromagnetic wave,

$\text{c}=\frac{1}{\sqrt{\mu_0\in_0}}=\frac{3\times10^8\text{m}}{\text{s}}$ 

which is constant

So It is independent of amplitude of electromagnetic wave, frequency and wavelength of electromagnetic wave.

so none of the above is correct.

1. $\hat{\text{k}}\text{ and }\hat{\text{j}}$

Explanation:

Electromagnetic wave is a transverse wave that means the electric and magnetic field associated to it will not only be perpendicular to each other but will also be

perpendicular to the direction in which the wave travels.

So, if waves travel along $\hat{\text{k}}$ direction then the electric and the magnetic field will be along $\hat{\text{i}}$ and $\hat{\text{j}}$​ directions.

1. AC only

Explanation:

For a capacitor, we have:

Q = CV

If Q is changing, there will be a current in capacitor plates,

$\text{I}=\frac{\text{dQ}}{\text{dt}}=\frac{\text{CdV}}{\text{dt}}$when voltage across the capacitor is constant, $\frac{\text{dV}}{\text{dt}}=0$

therefore, I = 0

It implies that, for a DC (constant) voltage, the capacitor current is zero.

Hence, for a DC source the conduction current and displacement current (capacitor current) are not same.

Whereas, by Maxwell's equation for a time varying voltage (AC voltage), both conduction and displacement currents are same.

1. 0.3m

Explanation:

Electromagnetic wave V = 3×10⁸m/s

Given frequency (f) = 10⁹Hz

$\text{V}=\text{f}\lambda$

$\lambda=\frac{\text{V}}{\text{f}}$

$=\frac{3 \times 10^8} {10^9}$

​= 0.3m