Wave Motion

21. If the potential difference applied to an X-ray tube is doubled while

If the potential difference applied to an X-ray tube is doubled while keeping the separation between the filament and the target as same, what will happen to the cutoff wavelength?

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Will remain same
Will be doubled
Will be halved
Will be four times of the original wavelength
This question was previously asked in
UPSC NDA-1 – 2017
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The question asks what happens to the cutoff wavelength of X-rays produced in an X-ray tube if the potential difference is doubled while keeping the filament-target separation the same.
The production of X-rays in an X-ray tube involves accelerating electrons through a potential difference ($V$) and making them strike a target. The energy gained by an electron accelerated through potential $V$ is $eV$, where $e$ is the elementary charge. This energy is converted into electromagnetic radiation (X-rays) and heat. The minimum wavelength ($\lambda_{min}$) of the emitted X-rays, also known as the cutoff wavelength or Duane-Hunt limit, occurs when the entire energy of the electron is converted into a single X-ray photon. The energy of a photon is $hf = hc/\lambda$, where $h$ is Planck’s constant, $f$ is frequency, $c$ is the speed of light, and $\lambda$ is the wavelength.
According to the Duane-Hunt law: $eV = hc/\lambda_{min}$
This equation shows that the cutoff wavelength $\lambda_{min}$ is inversely proportional to the applied potential difference $V$: $\lambda_{min} \propto 1/V$.
If the potential difference is doubled, i.e., $V’ = 2V$, the new cutoff wavelength $\lambda’_{min}$ will be:
$\lambda’_{min} = \frac{hc}{e(2V)} = \frac{1}{2} \left(\frac{hc}{eV}\right) = \frac{1}{2} \lambda_{min}$
So, the cutoff wavelength will be halved. The separation between the filament and the target does not directly affect the cutoff wavelength, which is determined by the maximum energy of the electrons reaching the target, dictated by the potential difference.
The X-ray spectrum produced consists of a continuous spectrum (bremsstrahlung) with a minimum wavelength $\lambda_{min}$ and characteristic peaks at specific wavelengths. The cutoff wavelength depends only on the accelerating voltage, while the intensity and characteristic peaks also depend on the target material and the electron beam current.

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22. Which one of the following is the correct relation between frequency $

Which one of the following is the correct relation between frequency $f$ and angular frequency $\omega$?

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$f = piomega$
$omega = 2pi f$
$f = 2omega/pi$
$f = 2piomega$
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UPSC NDA-1 – 2017
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The question asks for the correct relation between frequency ($f$) and angular frequency ($\omega$).
Frequency ($f$) represents the number of cycles or revolutions per unit of time (usually per second), measured in Hertz (Hz).
Angular frequency ($\omega$) represents the rate of rotation or oscillation in terms of angle per unit of time (usually radians per second), measured in rad/s.
One complete cycle or revolution corresponds to an angle of $2\pi$ radians. If $f$ cycles occur per second, then the total angle covered per second is $f \times (2\pi)$ radians.
Therefore, the angular frequency $\omega$ is related to the frequency $f$ by the equation:
$\omega = 2\pi f$
Angular frequency is commonly used in physics and engineering, particularly when dealing with circular motion, oscillations (like simple harmonic motion), and waves. It simplifies many equations compared to using frequency directly, especially in the context of derivatives and integrals involving sinusoidal functions. The unit of angular frequency is radians per second (rad/s), whereas the unit of frequency is Hertz (Hz) or cycles per second.

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23. Which one of the following sequences of electromagnetic spectrum is co

Which one of the following sequences of electromagnetic spectrum is correct in terms of ascending wavelength order ?

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X-rays – Gamma rays – UV – Infrared – Visible – Microwaves – Radio and TV waves
Gamma rays – X-rays – UV – Infrared – Visible – Microwaves – Radio and TV waves
Gamma rays – X-rays – UV – Visible – Infrared – Microwaves – Radio and TV waves
UV – Gamma rays – X-rays – Visible – Infrared – Microwaves – Radio and TV waves
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The electromagnetic spectrum is the range of all types of electromagnetic radiation. These radiations differ in wavelength, frequency, and energy. The question asks for the order of regions of the spectrum in terms of ascending (increasing) wavelength.
The order of the electromagnetic spectrum from shortest wavelength (highest frequency/energy) to longest wavelength (lowest frequency/energy) is:
Gamma rays → X-rays → Ultraviolet (UV) → Visible light → Infrared (IR) → Microwaves → Radio waves.
We need to arrange the given options in increasing wavelength.
Let’s check option C: Gamma rays – X-rays – UV – Visible – Infrared – Microwaves – Radio and TV waves. This sequence correctly lists the regions from shortest wavelength (Gamma rays) to longest wavelength (Radio waves). TV waves are a type of radio wave.
Comparing this with the other options, none of them follow the correct order of increasing wavelength.

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24. At a given point in space, the electric field associated with an elect

At a given point in space, the electric field associated with an electromagnetic wave is given by $\vec{E} = (2\hat{i} – 1.5\hat{j})e^{i[k_0(3x+4y)-\omega t]}$. At the same point, which one among the following is the correct value of the unit vector $(\hat{B})$ of the magnetic field associated with this electromagnetic wave ?

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$-hat{k}$
$1.5hat{i} + 2hat{j}$
$1.5hat{i} - 2hat{j}$
$3hat{i} - 4hat{j}$
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The correct value for the unit vector of the magnetic field is $-\hat{k}$.
For a plane electromagnetic wave propagating in a dielectric medium, the electric field vector ($\vec{E}$), the magnetic field vector ($\vec{B}$), and the wave vector ($\vec{k}$) are mutually perpendicular. The direction of wave propagation is given by the direction of $\vec{E} \times \vec{B}$, which is the same as the direction of $\vec{k}$.
The given electric field is $\vec{E} = (2\hat{i} – 1.5\hat{j})e^{i[k_0(3x+4y)-\omega t]}$. The term $k_0(3x+4y)$ represents $\vec{k} \cdot \vec{r}$. This indicates that the wave vector $\vec{k}$ is in the direction $3\hat{i} + 4\hat{j}$.
The direction of $\vec{E}$ is given by the amplitude vector $(2\hat{i} – 1.5\hat{j})$. Let this be $\vec{E}_0$.
We need to find a unit vector $\hat{B}$ such that $\vec{E}_0 \times \hat{B}$ is in the direction of $3\hat{i} + 4\hat{j}$.
Let’s check option A: $\hat{B} = -\hat{k}$.
$\vec{E}_0 \times \hat{B} = (2\hat{i} – 1.5\hat{j}) \times (-\hat{k}) = (2\hat{i} \times -\hat{k}) + (-1.5\hat{j} \times -\hat{k}) = 2\hat{j} + 1.5\hat{i} = 1.5\hat{i} + 2\hat{j}$.
The direction of this vector $1.5\hat{i} + 2\hat{j}$ is indeed the same as $3\hat{i} + 4\hat{j}$ (since $1.5\hat{i} + 2\hat{j} = \frac{1}{2}(3\hat{i} + 4\hat{j}) \times 2 = 3\hat{i} + 4\hat{j}$).
Thus, with $\hat{B} = -\hat{k}$, $\vec{E}_0 \times \hat{B}$ is in the correct direction of wave propagation.
Also, check perpendicularity: $\vec{E}_0 \cdot \hat{B} = (2\hat{i} – 1.5\hat{j}) \cdot (-\hat{k}) = 0$ and $(3\hat{i} + 4\hat{j}) \cdot (-\hat{k}) = 0$, confirming $\vec{E}$ and $\vec{k}$ are perpendicular to $\vec{B}$.
Options B, C, and D are not unit vectors, and therefore cannot be the unit vector $\hat{B}$.
The ratio of the magnitudes of the electric and magnetic fields in vacuum is constant, $|E|/|B| = c$, where c is the speed of light. In a medium, $|E|/|B| = v$, where v is the speed of light in the medium. The directions are related by $\vec{E} \times \vec{B} = v \mu \epsilon \vec{E} \times \vec{E} = v \vec{k}$ (assuming $\vec{E}$ and $\vec{B}$ are in phase). More accurately, the direction of propagation is $\vec{E} \times \vec{B}$.

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25. Which one of the following statements about X-rays is not correct?

Which one of the following statements about X-rays is not correct?

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X-rays are longitudinal waves.
X-rays are transverse waves.
X-rays are electromagnetic waves.
X-rays do not require a medium to propagate.
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UPSC Geoscientist – 2023
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X-rays are a form of electromagnetic radiation. Electromagnetic waves consist of oscillating electric and magnetic fields that are perpendicular to each other and to the direction of wave propagation. This characteristic defines them as transverse waves. Longitudinal waves, such as sound waves, are waves in which the oscillations are parallel to the direction of wave propagation. Therefore, the statement that X-rays are longitudinal waves is incorrect.
– X-rays are electromagnetic waves.
– Electromagnetic waves are transverse waves.
– Transverse waves have oscillations perpendicular to the direction of propagation.
– Longitudinal waves have oscillations parallel to the direction of propagation.
– X-rays, being electromagnetic, do not require a medium to travel.
The electromagnetic spectrum includes, in order of increasing frequency (decreasing wavelength): radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays. All waves in the electromagnetic spectrum share the fundamental property of being transverse waves and propagating at the speed of light in a vacuum.

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26. The characteristics of gravitational waves that make them difficult to

The characteristics of gravitational waves that make them difficult to detect are

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long wavelength and high energy
long wavelength and low energy
short wavelength and high energy
short wavelength and low energy
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Gravitational waves are difficult to detect primarily because they interact very weakly with matter, resulting in incredibly small amplitudes (strains) by the time they reach detectors on Earth. This weak interaction is associated with low energy flux at large distances from the source. The relevant wavelengths for detectable sources are often quite long (hundreds to thousands of kilometers for ground-based detectors, much longer for proposed space-based detectors), which necessitates very large and sensitive instruments like interferometers.
The difficulty in detecting gravitational waves stems from their weak coupling to matter. This means they cause only tiny distortions (strains) in spacetime as they pass through. Even waves from catastrophic events like black hole mergers result in strains of only about 10⁻²¹ to 10⁻²². Detecting such minuscule changes requires extraordinarily sensitive instruments, isolated from environmental noise. The energy carried by the waves, while immense near the source, spreads out over vast cosmic distances, leading to extremely low energy flux at the detector.
Ground-based detectors like LIGO and Virgo use interferometry to measure these minute changes in length caused by a passing gravitational wave. Future space-based detectors like LISA are designed to detect lower-frequency gravitational waves with much longer wavelengths, originating from different types of sources. The “long wavelength and low energy (flux at detection)” combination accurately reflects the challenges in detecting these elusive ripples in spacetime.

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27. Which one of the following statements, with regard to the properties o

Which one of the following statements, with regard to the properties of X-ray, is correct?

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X-rays can be deflected by an electric field
X-rays can be deflected by a magnetic field
X-rays are electromagnetic waves
X-rays are longitudinal waves
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UPSC Geoscientist – 2021
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X-rays are a form of electromagnetic radiation. This means they are waves composed of oscillating electric and magnetic fields that propagate through space.
Electromagnetic waves are not deflected by static electric or magnetic fields because they carry no net electric charge. They travel at the speed of light in a vacuum and are transverse waves (the oscillations of the electric and magnetic fields are perpendicular to the direction of propagation).
Other examples of electromagnetic waves include radio waves, microwaves, infra-red radiation, visible light, ultra-violet radiation, and gamma rays. Longitudinal waves, like sound waves, require a medium to travel and the oscillations are parallel to the direction of propagation.

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28. Which of the following statements is NOT true regarding the infrared

Which of the following statements is NOT true regarding the infrared radiation?

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It is electromagnetic in nature
Its wavelength is larger than that of the visible light
It can travel through vacuum
It is a longitudinal wave
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UPSC Geoscientist – 2020
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The statement that is NOT true regarding infrared radiation is that it is a longitudinal wave. Infrared radiation is a part of the electromagnetic spectrum. All electromagnetic waves, including visible light, radio waves, UV rays, X-rays, gamma rays, and infrared radiation, are transverse waves. In transverse waves, the oscillations (of electric and magnetic fields) are perpendicular to the direction of wave propagation. Longitudinal waves, such as sound waves in air, involve oscillations parallel to the direction of propagation.
Electromagnetic waves are transverse waves.
Infrared radiation is electromagnetic in nature, meaning it consists of oscillating electric and magnetic fields and travels at the speed of light in vacuum. Its wavelength range is typically longer than that of visible light (from about 700 nanometers to 1 millimeter). Like all electromagnetic waves, it can travel through vacuum. Infrared radiation is commonly associated with heat transfer (thermal radiation).

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29. X-rays can be used to : inspect welded joints between two metal part

X-rays can be used to :

  • inspect welded joints between two metal parts of a machine.
  • study structure of crystals.

Select the answer using the code given below :

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1 only
2 only
Both 1 and 2
Neither 1 nor 2
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UPSC CDS-2 – 2024
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Both inspecting welded joints between two metal parts of a machine and studying the structure of crystals can be done using X-rays.
X-ray radiography is a well-established Non-Destructive Testing (NDT) technique used to inspect internal structures for defects, such as detecting cracks, voids, or porosity in welded joints or castings. X-ray Diffraction (XRD) is a fundamental technique in materials science used to study the atomic and molecular structure of crystalline solids by analyzing the diffraction pattern produced when X-rays interact with the crystal lattice.
X-rays have numerous other applications, including medical imaging (radiography, CT scans), security screening, and X-ray fluorescence (XRF) spectroscopy for elemental analysis.

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30. Which one of the following statements with regard to the ultraviolet l

Which one of the following statements with regard to the ultraviolet light is not correct?

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It is an electromagnetic wave.
It can travel through vacuum.
It is a longitudinal wave.
Its wavelength is shorter/smaller than that of visible light.
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UPSC CDS-2 – 2020
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The statement that ultraviolet light is a longitudinal wave is not correct.
Ultraviolet (UV) light is a form of electromagnetic radiation, just like visible light, radio waves, and X-rays. All electromagnetic waves are transverse waves, meaning their oscillations are perpendicular to the direction of energy transfer (propagation). They can travel through vacuum at the speed of light. Longitudinal waves, such as sound waves, involve oscillations parallel to the direction of propagation and require a medium to travel. UV light has a shorter wavelength than visible light but a longer wavelength than X-rays and gamma rays.
UV radiation is invisible to the human eye but can cause effects like sunburn and skin cancer. It has various applications, including sterilization, medical therapy, and fluorescence. The electromagnetic spectrum is ordered by wavelength and frequency.

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