Q.Which part of the electromagnetic spectrum is used in RADAR ? Give its frequency range.

Showing the 12 most recent of 74 on this concept.

1. CBSE 2026Set 55/1/11 markMCQ
   Q.Electromagnetic waves used in a diagnostic tool in medicine have a wavelength range (A) $1\ \text{nm to } 10^{-3}\ \text{nm}$ (B) $400\ \text{nm to } 1\ \text{nm}$ (C) $1\ \text{mm to } 700\ \text{nm}$ (D) $0.1\ \text{m to } 1\ \text{mm}$

   ›Reveal solutionSolution

   The question asks for the wavelength range of electromagnetic waves used in a medical diagnostic tool. X-rays, used in radiography and CT scans, have wavelengths from about $1\ \text{nm}$ down to $10^{-3}\ \text{nm}$. The correct option is (A).

   The key here is to connect the medical diagnostic tool to the specific type of electromagnetic wave it uses. The most common diagnostic tools in medicine that rely on electromagnetic waves are X-ray machines (for bone fractures, chest X-rays, CT scans) and MRI machines (which use radio waves). The question gives wavelength ranges, so we need to match the correct range to the tool.

   X-rays are the standard for imaging hard tissues like bones. Their wavelengths are extremely short — much shorter than visible light. Visible light ranges from about $700\ \text{nm}$ (red) to $400\ \text{nm}$ (violet). X-rays lie beyond ultraviolet, in the range of roughly $1\ \text{nm}$ down to $0.001\ \text{nm}$ ($10^{-3}\ \text{nm}$). This matches option (A) exactly.

   Let’s check the other options to be sure:

   1. Option (B): $400\ \text{nm to } 1\ \text{nm}$ — This covers the ultraviolet and part of the visible spectrum. Ultraviolet is used in sterilisation, not in routine diagnostic imaging. So this is incorrect.

   2. Option (C): $1\ \text{mm to } 700\ \text{nm}$ — This spans the infrared region. Infrared is used in thermal imaging (thermography), but that is not a primary diagnostic tool like X-rays. The question likely refers to the most common tool, so this is not the best answer. …

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2. CBSE 2026Set 55/3/11 markMCQ
   Q.A plane electromagnetic wave travels through a medium and the magnetic field associated with it is given by $B = 5\times10^{-8}\,\sin(3\times10^{10}\,t - 150\,x)\,\text{T}$, where $x$ is in metres and $t$ is in seconds. The velocity of the wave is : (A) $2.0\times10^{8}\ \text{ms}^{-1}$ (B) $4.5\times10^{7}\ \text{ms}^{-1}$ (C) $3.5\times10^{7}\ \text{ms}^{-1}$ (D) $2.5\times10^{8}\ \text{ms}^{-1}$

   ›Reveal solutionSolution

   The wave velocity is found from the ratio $\omega/k$ in the given sinusoidal form. Here $\omega = 3\times10^{10}$ rad/s and $k = 150$ rad/m, giving $v = \omega/k = 2.0\times10^{8}$ m/s. The correct option is (A).

   The magnetic field is given as $B = 5\times10^{-8}\,\sin(3\times10^{10}\,t - 150\,x)$ T. This is a standard travelling wave expression of the form $B = B_0 \sin(\omega t - kx)$, where $\omega$ is the angular frequency and $k$ is the wave number. For any wave, the phase velocity is $v = \omega/k$. That is the direct route — no need to involve permittivity, permeability, or refractive index unless the medium is specified differently. Here the medium is simply given by the wave parameters themselves.

   1. Identify $\omega$ and $k$ from the equation.

      The term multiplying $t$ is $\omega = 3\times10^{10}$ rad/s.

      The term multiplying $x$ is $k = 150$ rad/m.

   2. Apply the wave velocity formula:

   $v = \frac{\omega}{k} = \frac{3\times10^{10}}{150}$

   3. Simplify: $v = \frac{3\times10^{10}}{1.5\times10^{2}} = 2\times10^{8}\ \text{m/s}$ …

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3. CBSE 2026Set ANNUAL1 mark
   Q.Which electromagnetic wave has the highest energy?

   ›Reveal solutionSolution

   Energy of an EM wave is proportional to its frequency; gamma rays have the highest frequency.

   Photon energy $E=hf$, so the electromagnetic wave with the highest frequency has the highest energy. In the electromagnetic spectrum, gamma …

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4. CBSE 2026Set DS1 markMCQ
   Q.Electromagnetic waves of minimum frequency is:

   i) ultraviolet rays

   ii) $X$ rays

   iii) $\gamma$-rays

   iv) microwaves

   ›Reveal solutionSolution

   Microwaves have the longest wavelength (lowest frequency) of the four, so they are the minimum-frequency wave.

   Concept. In the electromagnetic spectrum, frequency increases (and wavelength decreases) in the order:

   $\text{radio} < \text{microwave} < \text{infrared} < \text{visible} < \text{ultraviolet} < \text{X-rays} < \gamma\text{-rays}.$ …

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5. CBSE 2026Set A1 markMCQ
   Q.The dimensions of B0^2/μ0 will be the same as that of (A) energy density (B) work (C) momentum (D) electric flux

   ›Reveal solutionSolution

   Magnetic energy density = B²/2μ₀, so B²/μ₀ carries the dimensions of energy density.

   The energy stored per unit volume in a magnetic field is $u_B = \dfrac{B^2}{2\mu_0}$.

   Apart from the numerical factor ½, the quantity $\dfrac{B^2}{\mu_0}$ therefore has the dimensions of energy density (J/m³).

   …

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6. CBSE 2026Set ANNUAL1 markMCQ
   Q.Which of the following has the smallest wavelength?

   (a) Micro-waves

   (b) Red light

   (c) Ultraviolet radiation

   (d) gamma-rays

   ›Reveal solutionSolution

   Across the electromagnetic spectrum, wavelength decreases (and frequency/energy increases) in the order: micro-waves > red light (visible) > ultraviolet > gamma-rays.

   The electromagnetic spectrum, arranged from LONGEST to SHORTEST wavelength, runs roughly: radio waves > microwaves > infrared > visible light (red to violet) > ultraviolet > X-rays > gamma rays. Comparing the four given options in that order - micro-waves (longest here), red light (visible, shorter than microwaves), …

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7. CBSE 2026Set ANNUAL1 markMCQ
   Q.Identify the constituent radiation of electromagnetic spectrum which is used in medicine to destroy cancer cells.

   (a) X-rays

   (b) UV rays

   (c) IR rays

   (d) Gamma rays

   ›Reveal solutionSolution

   Gamma rays, the highest-energy, shortest-wavelength part of the EM spectrum, are used in radiotherapy to destroy cancer cells.

   Among the options, X-rays are chiefly used for imaging (and in some radiotherapy), UV rays for sterilisation and detecting fluorescence, and IR rays for heating and thermal imaging. Gamma rays, emitted by radioactive nuclei, carry very high photon energy and deep penetrating power. In medicine this is exploited in gamma-ray therapy (e.g. Cobalt-60 teletherapy, the 'gamm …

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8. CBSE 2026Set ANNUAL1 markMCQ
   Q.The velocity of electromagnetic waves in vacuum is:

   (a) c = 1/√(μ₀ε₀)

   (b) c = √(μ₀ε₀)

   (c) c = √(μ₀/ε₀)

   (d) c = √(ε₀/μ₀)

   ›Reveal solutionSolution

   Maxwell's equations predict electromagnetic waves travelling in vacuum at speed $c=1/\sqrt{\mu_0\varepsilon_0}$, which numerically matches the measured speed of light.

   Starting from Maxwell's equations in free space (no charges or currents), the wave equations for $\vec E$ and $\vec B$ both take the standard wave-equation form with wave speed $v=1/\sqrt{\mu_0\varepsilon_0}$. Substituting $\mu_0=4\pi\times10^{-7}\ \text{T m/A}$ and $\varepsilon_0=8.85\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$ gives $v\approx3\times10^{8}\ \text{m/s}$, exactly the speed of light — this a …

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9. CBSE 2026Set ANNUAL1 markMCQ
   Q.In a plane electromagnetic wave the electric field oscillates sinusoidally with a frequency $2.5\times10^{10}Hz$ and amplitude $480V/m$. The amplitude of the oscillating magnetic field will be –

   (a) $1.52\times10^{-8}\ weber/m^2$

   (b) $1.52\times10^{-7}\ weber/m^2$

   (c) $1.6\times10^{-6}\ weber/m^2$

   (d) $1.6\times10^{-7}\ weber/m^2$

   ›Reveal solutionSolution

   $B_0 = E_0/c$.

   In a plane electromagnetic wave, the amplitudes of the electric and magnetic fields are related by $B_0 = \dfrac{E_0}{c}$:

   …

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10. CBSE 2025Set 55/4/11 markMCQ
    Q.The amplitude of the electric field in an electromagnetic wave in free space is $1000$ Vm$^{-1}$. The amplitude of the magnetic field in this electromagnetic wave is: (A) $3.0 \times 10^{-3}$ T (B) $3.33 \times 10^{-8}$ T (C) $3.0 \times 10^{11}$ T (D) $3.33 \times 10^{-6}$ T

    ›Reveal solutionSolution

    In an electromagnetic wave, the electric and magnetic field amplitudes are related by the speed of light: $E_0 = c B_0$. With $E_0 = 1000$ V/m, we find $B_0 = 3.33 \times 10^{-6}$ T.

    Why the fields are linked by $c$

    An electromagnetic wave is a self-sustaining oscillation of electric and magnetic fields that propagate together through space. Maxwell's equations demand a precise relationship between these two fields: at every instant and every point in the wave, the ratio of the electric field amplitude to the magnetic field amplitude equals the speed of light in that medium.

    In free space, this relationship is beautifully simple:

    $E_0 = c B_0$

    where $E_0$ is the amplitude of the electric field, $B_0$ is the amplitude of the magnetic field, and $c = 3 \times 10^8$ m/s is the speed of light in vacuum. This isn't arbitrary—it emerges directly from the wave equations derived from Maxwell's laws. The electric and magnetic fields are perpendicular to each other and to the direction of propagation, oscillating in phase, with their amplitudes locked in this ratio.

    Finding the magnetic field amplitude

    We're given the electric field amplitude and need to find the magnetic field amplitude.

    1. Write down the fundamental relationship:

    $E_0 = c B_0$

    2. Rearrange to solve for $B_0$:

    $B_0 = \frac{E_0}{c}$

    3. Substitute the given values: …

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11. CBSE 2025Set 55/6/11 markMCQ
    Q.Assertion (A): X-rays are produced when slow moving electrons are stopped by a metal target of high atomic number. Reason (R): X-rays consist of low-energy photons. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Both Assertion (A) and Reason (R) are false.

    ›Reveal solutionSolution

    Both statements are false: X-rays require FAST (not slow) electrons to be decelerated by a high-Z target, and X-ray photons are HIGH-energy (not low-energy). The correct option is (D).

    Concept and intuition

    X-rays are electromagnetic radiation of very short wavelength (roughly $10^{-11}$ m to $10^{-8}$ m) and correspondingly very HIGH photon energy — from about $100\,\text{eV}$ up to $100\,\text{keV}$, thousands of times more energetic than visible-light photons ($\sim2$–$3\,\text{eV}$). They are produced in an X-ray tube by bremsstrahlung ("braking radiation"): electrons are first accelerated through a large potential difference (tens of kilovolts), giving them very HIGH kinetic energy, and are then suddenly decelerated on striking a metal target of high atomic number (e.g. tungsten, $Z=74$) — a high-$Z$ target is used because the efficiency of bremsstrahlung production increases with $Z$.

    Evaluating the Assertion

    The Assertion states that X-rays are produced when slow moving electrons are stopped by a high-$Z$ target. This gets the mechanism backwards: it is precisely because the electrons are moving very fast (having been accelerated through a large voltage) that stopping them suddenly releases enough energy to produce X-ray photons. A genuinely slow-moving electron carries far too little kinetic energy to produce a photon anywhere in the X-ray range. So the Assertion, as literally worded, is false — the single word "slow" (which should read "fast") is what makes it false; the rest of the statement (high-Z target, sudden stopping) is correct in isolation, but the assertion as a whole is not.

    Evaluating the Reason

    The Reason states that X-rays consist of low-energy photons. This is also the opposite of the truth: X-ray photons are HIGH-energy compared to visible, infrared, or radio-wave photons — that is exactly why they can penetrate soft tissue and are dangerous in large doses. So the Reason is false as well. …

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12. CBSE 2025Set ANNUAL1 markMCQ
    Q.A plane electromagnetic wave travels in free space along x-direction. At a point in space and time, $\vec{E} = 9.0\hat{j}$ V/m. Magnitude of $\vec{B}$ at this point is-

    (a) $3\times10^{-8}$ T

    (b) $9\times10^{-8}$ T

    (c) $27\times10^{9}$ T

    (d) $2.1\times10^{-8}$ T

    ›Reveal solutionSolution

    In a plane EM wave in free space, $B = E/c$.

    For a plane electromagnetic wave travelling in free space, the magnitudes of the electric and magnetic fields are related by

    $B = \frac{E}{c}$ …

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