Physics

Electromagnetic Waves and Spectrum

670 Questions

Electromagnetic waves and spectrum questions test a candidate's understanding of radiation frequencies, wavelengths, and the properties of different rays like infrared, ultraviolet, and visible light. Concepts also cover practical applications in astronomy and the fundamental speed of light calculations. This physics topic appears regularly in general science sections of major competitive examinations.

UV rays propertiesElectromagnetic radiation frequencyWavelength identificationSpeed of light calculationsBlack body radiation

Electromagnetic Waves and Spectrum Questions

Multiple choice physics measurements and experimentation measuring distance of celestial bodies unconventional units of measurements units of mass

1 light year is equal to 

  1. $\displaystyle 6.3\times { 10 }^{ 5 }\overset { o }{ A } $
  2. $\displaystyle 6.3\times { 10 }^{ 4 }$$AU$
  3. $\displaystyle 3.0\times { 10 }^{ 16 }{ ms }^{ -1 }$
  4. $\displaystyle 6.3\times { 10 }^{ 4 }\overset { o }{ A } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
We know that   $1 \ light year  = 9.46\times 10^{15} \ m$
Also,   $1 \ AU = 1.5\times 10^{11} \ m$
Thus   $1 \ light year  = \dfrac{9.46\times 10^{15}}{1.5\times 10^{11}}  \ AU= 6.3\times 10^4 \ AU$
Multiple choice polarisation of light polarisation wave optics optics physics

Which of the following cannot be polarised ?

  1. Radio waves
  2. $\beta$ rays

  3. Infrared rays
  4. $\gamma$ rays
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$ \beta $ are stream of particles comprising of electrons moving with very high velocity, hence it cannot be polarized, while others can as they are electromagnetic waves.

Multiple choice physics energy production perfectly black body black-body radiation black body radiation

The rate of emission of radiation of a black body at 273$^{ \circ  }{ C }$ is E, then the rate of emission of radiation  of this body at 0$^{ \circ  }{ C }$ will be :-

  1. $\dfrac { E }{ 16 } $
  2. $\dfrac { E }{ 4 } $
  3. $\dfrac { E }{ 8 } $

  4. 0
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Stefan-Boltzmann law states E is proportional to T^4. T1 = 273 + 273 = 546K. T2 = 0 + 273 = 273K. Ratio E2/E1 = (273/546)^4 = (1/2)^4 = 1/16. Thus E2 = E/16.

Multiple choice physics observing space: telescopes space research and satellites space travel space exploration and forms of light

The number of wavelengths in the visible spectrum is :

  1. $4000$
  2. $6000$
  3. $2000$
  4. $inifinite$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
  • The visible light spectrum is the section of the electromagnetic radiation spectrum that is visible to the human eye. 
  • It ranges in wavelength from approximately 400 nanometers ($4 \times  10 ^{-7}$ m, which is violet) to 700 nm ($7 \times 10^{-7}$ m, which is red).
  • Hence wavelengths are infinite
  • Option D is the right answer
Multiple choice physics observing space: telescopes space research and satellites space travel space exploration and forms of light

Choose the correct answer from the alternatives given.
Which of the following ray are not electromagnetic waves

  1. X-rays
  2. $\gamma- $ rays
  3. $\beta -$ rays

  4. Heat rays
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Cosmic rays, γ-rays, and X-rays are part of electromagnetic spectrum, while β-rays are emitted by radioactive elements. Hence β-rays is not electromagnetic waves.

Multiple choice physics units and measurement: error analysis rounding off digits rounding of digits standard form

What is the order of magnitude of one light year? 

  1. $10^{15} m$
  2. $10^{10} m$
  3. $9.2 \times 10^{15} m$
  4. $10^{16} m$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

One light year is the distance of light at a speed $3\times 10^8 m/s$ in 1 year. 
Thus, $1 $ light year $= 3\times 10^8 m/s\times 1$ year $=3\times 10^8 m/s\times (365\times 24\times 3600 s)=9.5\times 10^{15} m$
According to rule of order of magnitude, as $9.5>5$ so 9.5 will be taken as 10.
Thus, order of magnitude of one light year $=10\times 10^{15}=10^{16} m$

Multiple choice physics behaviour of perfect gas and kinetic theory of gases degree of freedom: law of equipartition of energy law of equipartition of energy law of equipartition of energy and mean free path

The de-Broglie wavelength of a particle accelerated with $150\ volt$ potential is $10^{-10}\ m$. If it accelerated by $600\ volts$ p.d. its wavelength will be

  1. $0.25\ A^{o}$
  2. $0. 5\ A^{o}$
  3. $1.5\ A^{o}$
  4. $2\ A^{o}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given,

$\lambda =\dfrac{hc}{eV}\ \ \ \ where,\ V=potential$

$\lambda \ \alpha \ \dfrac{1}{V}$

${{10}^{-10}}\ \alpha \ \dfrac{1}{150}\ ......\ (1)$

$\lambda \ \alpha \ \dfrac{1}{600}\ ......\ (2)$

Divide (2) by (1)

$ \dfrac{\lambda }{{{10}^{-10}}}=\dfrac{150}{600}=\dfrac{1}{4} $

$ \Rightarrow \lambda =0.25\times {{10}^{-10}}m\ =0.25\ {{A}^{o}} $ 

Multiple choice stefan's law black body radiation heat transfer thermal properties physics

The radiation emitted by a star $A$ is $10000$ times that of the sun. If the surface temperature of the sun and star $A$ are $6000:K$ and $2000:K$, respectively, the ratio of the radii of the star $A$ and the sun is

  1. $300:1$
  2. $600:1$
  3. $900:1$
  4. $1200:1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Stars can be approximated as black bodies.
Hence by stefan's law, power emitted=P=$\sigma AT^4$
Thus,$ \dfrac{P _A}{P _{sun}}=\dfrac{r _A^2T _A^4}{r _{sun}^2T _{sun}^4}$
$\Rightarrow (\dfrac{r _A}{r _{sun}})^2=10000\times (\dfrac{6000}{2000})^4 \rightarrow \dfrac{r _A}{r _{sun}}=900$
So required ratio is 900:1
Multiple choice stefan's law black body radiation heat transfer thermal properties physics

The radiation emitted by a perfectly black body is proportional to 

  1. temperature on ideal gas scale

  2. fourth root of temperature on ideal gas scale

  3. fourth power of temperature on ideal gas scale

  4. square of temperature on ideal gas scale
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Stefan-Boltzmann law states that total power radiated by a perfectly black body is
$P=A\sigma { T }^{ 4 }$
so the radiation emitted by a perfectly black body is proportional to fourth power of temperature on ideal gas scale.
option (C) is the correct answer.

Multiple choice stefan's law black body radiation heat transfer thermal properties physics

The rate of radiation from a black body at $0$$^{o}$C is $E$. The rate of radiation from this black body at $273$$^{o}$C is :

  1. $2E$
  2. $E/2$
  3. $16E$
  4. $E/16$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

As we know that: ${E} \ {\propto} \ {T}^{4}$
So, $\dfrac{E}{{T} _{1}^{4}}=\dfrac{{E} _{2}}{{T} _{2}^{4}}$
${E} _{2}=\dfrac{{T} _{2}^{4}\times{E}}{{T} _{1}^{4}}=\dfrac{{546}^{4}}{{273}^{4}}$
${E} _{2}={16E}$

Multiple choice stefan's law black body radiation heat transfer thermal properties physics

If a graph is plotted by taking spectral emissive power along $y-$axis and wavelength along x-axis is:

  1. Emissivity

  2. Total intensity of radiation

  3. Diffusivity

  4. Solar constant
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Emissive power varies according to Stefan-Boltzmann law as :

$E=\sigma {{T}^{4}}$

According to Planck’s distribution law:

${{E} _{\lambda }}(\lambda ,T)=\dfrac{{{C} _{1}}}{{{\lambda }^{5}}\left[ \exp (\dfrac{{{C} _{2}}}{\lambda T})-1 \right]}$

The graph shows that the emitted radiation varies with wavelength and also it shows Emissivity.