Properties of EM Waves - NEET Physics Chapterwise MCQs & PYQs
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NEET Properties of EM Waves MCQs & PYQs
Practice NEET Properties of EM Waves Questions
Question 11:
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.
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 $c = \frac{1}{\sqrt{\mu_0\epsilon_0}}$ .
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 $\bar{E} \times \bar{B}$ .
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 $\lambda = \frac{3 \times 10^8}{10^7} = 30 m$ .
If $\epsilon_0$ and $\mu_0$ are the electric permittivity and magnetic permeability in a free space, $\epsilon$ and $\mu$ are the corresponding quantities in medium, the refractive index of the medium is:
(1994)
The refractive index is defined as $n = \frac{c}{v}$ . We know $c = \frac{1}{\sqrt{\mu_0\epsilon_0}}$ and $v = \frac{1}{\sqrt{\mu\epsilon}}$ . Substituting these expressions, we get $n = \frac{1 / \sqrt{\mu_0\epsilon_0}}{1 / \sqrt{\mu\epsilon}} = \sqrt{\frac{\mu\epsilon}{\mu_0\epsilon_0}}$ .
The frequency of electromagnetic wave, which best suited to observe a particle of radius $3 \times 10^{-4} cm$ is of the order of
(1991)
To effectively observe a particle, the resolving electromagnetic wave must have a wavelength of the order of the particle's size: $\lambda \approx 3 \times 10^{-4} cm = 3 \times 10^{-6} m$ . The corresponding frequency is $f = \frac{c}{\lambda} = \frac{3 \times 10^8}{3 \times 10^{-6}} = 10^{14} Hz$ .
Choose the correct answer from the options given below:
(2022)
Based on the standard electromagnetic spectrum: AM radio waves have the longest wavelength ( $\sim 10^2 m$ ). Microwaves are shorter ( $\sim 10^{-2} m$ ). Infrared is shorter still ( $\sim 10^{-4} m$ ). X-rays have very short wavelengths ( $\sim 10^{-10} m$ ). This matches option a.
The E.M. wave with shortest wavelength among the following is:
(2020-Covid)
In the electromagnetic spectrum, gamma rays have the highest frequency and the most energy, which corresponds to the shortest wavelength among all the listed options.
Which colour of the light has the longest wavelength?
(2019)
In the visible light spectrum (VIBGYOR), the wavelength progressively increases from the violet end to the red end. Thus, red light has the longest wavelength (approximately $700 nm$ ).
The energy of the E.M. waves is of the order of $15 keV$ . To which part of the spectrum does it belong?
(2015 Pre)
The wavelength associated with $15 keV$ is $\lambda = \frac{hc}{E} = \frac{1240 eV\cdot nm}{15 \times 10^3 eV} \approx 0.08 nm$ . A wavelength of $0.08 nm$ (or $0.8 \mathring{A}$ ) falls squarely in the X-ray region of the electromagnetic spectrum.