NEET Properties of EM Waves MCQs & PYQs

Practice NEET Properties of EM Waves Questions

Question 31:

easy

Light 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 $F = \frac{2 P}{c}$.
$F = \frac{2 \times 375}{3 \times 10^8} = \frac{750}{3 \times 10^8} = 2.50 \times 10^{-6} N$.

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Question 32:

easy

The ratio of amplitude of magnetic field to the amplitude of electric field for an electromagnetic wave propagating in vacuum is equal to:

(2012 Mains)

The speed of an electromagnetic wave in vacuum is related to the field amplitudes by $c = \frac{E_0}{B_0}$ . Rearranging this, we get the ratio of magnetic to electric field amplitude as $\frac{B_0}{E_0} = \frac{1}{c}$ . Therefore, it is the reciprocal of the speed of light in vacuum.

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Question 33:

easy

The dimensions of $(\mu_0\epsilon_0)^{-\frac{1}{2}}$ are:

(2011 Pre)

The expression $(\mu_0\epsilon_0)^{-\frac{1}{2}} = \frac{1}{\sqrt{\mu_0\epsilon_0}}$ represents the speed of light in vacuum (c). The dimensional formula for velocity or speed is $[L T^{-1}]$ .

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Question 34:

easy

The electric and the magnetic field associated with an e.m. wave, propagating along the +z axis, can be represented by:

(2011 Pre)

The direction of propagation of an electromagnetic wave is parallel to the Poynting vector, given by the cross product $\bar{E} \times \bar{B}$ . For option a, $\hat{i} \times \hat{j} = \hat{k}$ , which corresponds to the +z axis.

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Question 35:

easy

The electric field of an electromagnetic wave in free space is given by $E = 10\cos(10^7 t + kx)\hat{j} V/m$ where t and x are in second and metres respectively. It can be inferred that

1. The wavelength $\lambda$ is $188.4 m$

2. The wave number k is $0.33 rad/m$

3. The wave amplitude is $10 V/m$

4. The wave is propagating along +x direction

Where one of the following pairs of statement is correct?

(2010 Mains)

Comparing with standard form $E = E_0\cos(\omega t + kx)$ , amplitude $E_0 = 10 V/m$ (Statement 3 is correct). Angular frequency $\omega = 10^7 rad/s$ , so wave number $k = \frac{\omega}{c} = \frac{10^7}{3 \times 10^8} = 0.033 rad/m$ . Wavelength $\lambda = \frac{2\pi}{k} = \frac{2\pi}{0.033} \approx 188.4 m$ (Statement 1 is correct). The positive sign indicates propagation in the -x direction. Thus, 1 and 3 are correct.

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Question 36:

easy

Which of the following statement is false for the properties of electromagnetic waves?

(2010 Pre)

In an electromagnetic wave, the oscillating electric and magnetic field vectors are mutually perpendicular to each other, not parallel. They are also perpendicular to the direction of propagation. Thus, option d is false.

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Question 37:

easy

The electric field part of an electromagnetic wave in a medium is represented by $E_x = 0$
$E_y = 2.5\cos\left[\left(2\pi \times 10^6 \frac{rad}{m}\right)t – \left(\pi \times 10^{-2} \frac{rad}{s}\right)x\right]$
$E_z = 0$ . The wave is:

(2009)

Comparing with $E_y = E_0\cos(\omega t - kx)$ , we find $\omega = 2\pi \times 10^6 rad/s$ and $k = \pi \times 10^{-2} rad/m$ . Frequency $f = \frac{\omega}{2\pi} = 10^6 Hz$ . Wavelength $\lambda = \frac{2\pi}{k} = \frac{2\pi}{\pi \times 10^{-2}} = 200 m$ . The negative sign between the temporal and spatial terms indicates it moves in the +x direction.

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Question 38:

easy

The velocity of electromagnetic radiation in a medium of permittivity $\epsilon_0$ and permeability $\mu_0$ is given by:

(2008)

According to Maxwell's theory of electromagnetism, the speed of electromagnetic waves in free space is determined by its electric and magnetic properties and is given by the formula $c = \frac{1}{\sqrt{\mu_0\epsilon_0}}$ .

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Question 39:

easy

The velocity of electromagnetic wave is parallel to:

(2002)

The direction of propagation of an electromagnetic wave, and thus its velocity vector, is given by the direction of the Poynting vector $\bar{S} = \frac{1}{\mu_0} (\bar{E} \times \bar{B})$ . Hence, it is parallel to $\bar{E} \times \bar{B}$ .

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Question 40:

easy

Frequency of an E.M. waves is $10 MHz$ then its wavelength is:

(1999)

Using the wave equation $c = f\lambda$ , we can solve for wavelength $\lambda = \frac{c}{f}$ . Given $f = 10 MHz = 10^7 Hz$ and $c = 3 \times 10^8 m/s$ , we have $\lambda = \frac{3 \times 10^8}{10^7} = 30 m$ .

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