Electromagnetic Waves MCQ Quiz in தமிழ் - Objective Question with Answer for Electromagnetic Waves - இலவச PDF ஐப் பதிவிறக்கவும்

Solution:

 the intensity expression:

\(\ I = \frac{1}{2} c \epsilon_0 E_0^2\)

Where:
c is the speed of light,
\(\epsilon_0\) is the permittivity of free space,
\(E_0\) is the electric field.

The power per unit area P/A is given by:

\(\ \frac{P}{A} = \frac{1}{2} c \epsilon_0 E_0^2\)

Therefore, for power P :

\(\ P = \frac{1}{2} c \epsilon_0 E_0^2 \times \text{Area}\)

Additionally, for the magnetic field B_0 , the power can also be written as:

\(\ P = \frac{1}{2} c \epsilon_0 B_0^2 c^2 \times \text{Area}\)

Substituting values, we proceed to compute the area:

\(\ \text{Area} = \frac{1}{2} \epsilon_0 c^3\)

Next, the area formula for a circular aperture is used:

\( \text{Area} = \frac{1}{2} A^2 \times \pi \times (2 \times 10^{-3})^2\)

Given:

\(\ P = 15.7 \times 10^{-3}\)

Solving for A :

\(\ 15.7 \times 10^{-3} = \frac{1}{2} A^2 \times \pi \times (2 \times 10^{-3})^2\)

Simplifying:

\(\ A = \sqrt{2500} = 50\)

Thus, the value of A is 50.

The correct option is (1) : 50 

Explanation:

When a light rays pass through one medium to another medium , the light gets reflected as well as refracted..

The snells law says that 

\(n_1 \sinθ_1=n_2\sinθ_2\)

The \(n_1 \text{ & } n_2\) are refractive index medium 1 and 2.

The θ₁ & θ₂ are the angle of incidence and refraction respectively.

Given :

θ₁ =θ _B 

n₁ =1.5

\(n_2= 2.0\)

The Brewster angle is angle at which refractive rays and reflected rays are perpendicular to each other.  Thus α =90^∘ .

Verification:

θ_B=\(\tan^{-1}(\frac{n_2}{n_1})=53.13^\circ\)

The angle of  refraction is \(\theta_2=\frac{n_1\sin\theta_B}{n_2}=36.87^\circ\)

From the below diagram,

α =\(180^\circ-(36.87+56.13)=90^\circ\)

The correct option is (3).

Explanation:

To determine the incorrect statement about electromagnetic waves, let's analyze each option:

The electromagnetic field vectors E and B are mutually perpendicular.

Correct – In an electromagnetic wave, the electric field (E) and the magnetic field (B) are perpendicular to each other and also to the direction of wave propagation.

The electromagnetic field vectors E and B are in the same phase.

Correct – The electric and magnetic fields in an electromagnetic wave oscillate in phase with each other. That means their maxima, minima, and zero crossings occur at the same time.

The electromagnetic waves are transverse.

Correct – Electromagnetic waves are transverse in nature, meaning both E and B fields oscillate perpendicular to the direction of wave propagation.

Thus, option '5' is correct.

The Correct Answer is 3 × 108 m/s.

Key Points

Electromagnetic waves

• It is a transverse wave defined as composed of oscillating electrical and magnetic fields perpendicular (90°) to each other.
• The Speed of electromagnetic waves is the same as of speed of light c = 3 × 10⁸ m/s. 
• Speed of electromagnetic waves is dependent upon medium as, v=nc​
• where C is the speed of light in vacuum, N is the refractive index in the medium.
• Speed is dependent on wavelength and frequency as v=λν
• It is independent of intensity.
• All electromagnetic waves travel at the same speed through empty space.
• That speed, called the speed of light, is about 300 million meters per second (3.0 x 10⁸ m/s). 
• The velocity of a wave of electromagnetic on every medium is the same for all intensities.

Concept:

According to Fresnel equations, fraction of incident energy that gets transmitted when light is incident normally is given by \(T=\frac{4\times n_t n_i}{(n_i+n_t)^2}\)

Here, \(n_t=\)refractive index of glass

\(n_i=\)refractive index of air

Explanation:

For air-glass interface, \(n_t=1.5\) and \(n_i=1\)

• \(T=\frac{4\times n_t n_i}{(n_i+n_t)^2}\)

• \(T=\frac{4\times 1.5\times 1}{(1.5+1)^2}=\frac{6\times10\times10}{25\times 25} =\frac{24}{25}\)

So, the correct answer is \(T=\frac{24}{25}\).

Concept:

 We are using Snell's law at the interface of dielectrics with respective permittivities at the normal,

• \(n_1sin\theta_1=n_2sin\theta_2\)
• \(n_1=\sqrt{\frac{1}{\mu_0 \epsilon_1}}\)
• where n is the refractive index and \(\epsilon\) is the permittivity of media and \(\theta\) is the angle with the normal at the interface of two dielectrics.

Explanation: 

• Using Snell's law: \(n_1sin\theta_1=n_2sin\theta_2\)

We will take \(\theta=\theta _b \) at the interface and \(\theta_1+\theta_2=90^0\)

\(n_1sin\theta_b=n_2sin(90-\theta_b)=n_2cos\theta_b\)

\(\frac {sin\theta_b}{cos\theta_b}=tan\theta_b=\frac {n_2}{n_1}\)

• For angles \(\theta_1\) and \(\theta_2\)

\(\frac {tan\theta_1}{tan\theta_2}=\frac{n_2^2}{n_1^2}\)

• Now, \(n_1=\sqrt{\frac{1}{\mu_0 \epsilon_1}}\) and \(n_2=\sqrt{\frac{1}{\mu_0 \epsilon_2}}\)
• Substitute, we get, \(\frac {tan\theta_1}{tan\theta_2}=\frac{n_2^2}{n_1^2}\)\(=\frac{\epsilon_1}{\epsilon_2}\)

​\(\frac {tan\theta_1}{tan\theta_2}\)\(=\frac{\epsilon_1}{\epsilon_2}\)

So, the correct answer is \(\frac{\epsilon_1}{\epsilon_2}\)\(=\frac {tan\theta_1}{tan\theta_2}\).

Concept:

 Here, according to Biot-Savart law for magnetic induction, 

• \(\oint B.dl=\mu_0I\)

Calculation-

Given that an infinitely long straight wire with distance d and current I is flowing along z-direction:

• \(\oint B\cdot dl=\mu_0I\)

The length element will be: \(dl=2\pi d\)

• \(B\times 2\pi d=\mu_0I\)
• \(B=\frac{\mu_0 I}{2\pi d}\)

Multiply and divide by 2, we get,

• \(B=(\frac{\mu_0}{4\pi})(\frac {2I}{d})\)

Now, in z direction, we take the unit vector and replace \(\mu_0\) from \(M_0\), we get,

• \(B=(\frac{M_0}{4\pi})(\frac {2I}{d})\hat k\)

So, the correct answer is \(B=(\frac{M_0}{4\pi})(\frac {2I}{d})\hat k\).