Physics

Electromagnetic Waves and Spectrum

659 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 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 mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

The electric field of an electromagnetic wave is given by, $E=(50N^{-1})\, \sin { \omega  } (t-x/c)$. Find the energy contained in a cylinder of cross section $10cm^2$ and length $50 cm$ along the x-axis.

  1. $5.5\times 10^{-12}J$
  2. $4.5\times 10^{-12}J$
  3. $5\times 10^{-13}J$
  4. $3.5\times 10^{-10}J$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The energy density of an EM wave is u = (1/2) * epsilon0 * E^2 + (1/2) * (B^2 / mu0). For an EM wave, the average energy density is u_avg = (1/2) * epsilon0 * E0^2. The total energy is U = u_avg * Volume. Volume = Area * length = 10 * 10^-4 m^2 * 0.5 m = 5 * 10^-4 m^3. E0 = 50 V/m. u_avg = 0.5 * 8.85 * 10^-12 * 50^2 = 1.1 * 10^-8 J/m^3. U = 1.1 * 10^-8 * 5 * 10^-4 = 5.5 * 10^-12 J.

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

The radiation emitted by a star $A$ is $1000$ times that of the sun. If the surface temperatures 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

$E\propto A{ T }^{ 4 }$


$\displaystyle \frac { { E } _{ 1 } }{ { E } _{ 2 } } =\frac { 1000 }{ 1 } =\frac { \pi { { r } _{ 1 } }^{ 2 }\times { T }^{ 4 } }{ \pi { { r } _{ 2 } }^{ 2 }\times { T }^{ 4 } } =\frac { { { r } _{ 1 } }^{ 2 }\times { T }^{ 4 } }{ { { r } _{ 2 } }^{ 2 }\times { T }^{ 4 } } =\frac { { { r } _{ 1 } }^{ 2 }\times { 2000 }^{ 4 } }{ { { r } _{ 2 } }^{ 2 }\times 6000^{ 4 } } =\frac { { { r } _{ 1 } }^{ 2 } }{ { { r } _{ 2 } }^{ 2 }\times 81 } $

$\displaystyle \frac { { r } _{ 1 } }{ { r } _{ 2 } } =\sqrt { \frac { 1000 }{ 1 } \times \frac { 81 }{ 1 }  } =284.6:1$

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

The amount of thermal radiations emitted from one square centimeter area of a black body in a second when at a temperature of 1000K

  1. 5.67 J

  2. 56.7 J

  3. 567 J

  4. 5670 J
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Stefan's law $\Delta Q = \sigma ST^4\Delta t$
$\sigma=5.67E- 8W/m^2K^4;\, S=1E-4m^2;\, T=1000K; \, \Delta t=1s;$
subsutituting value in Stefan's law $\Delta Q=5.67J$
The value can be directly calculated by the Stefan's equation. After substituting the parameters sigma $= 5.67 E-8  W/m^2/K^4, A=10^-4 m^4,$ $T=1000K$ the value comes $5.67$ J
Thus, A is correct answer.
Multiple choice stefan's law black body radiation heat transfer thermal properties physics

The rate of radiation of a black body at $0^{\circ}C$ is $E$ J/s. Then the rate of radiation of this black body at $273^{\circ}C$ will be

  1. 16 E

  2. 8 E

  3. 4 E

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

From the stefan's law: ${E}={\sigma}{A}{T}^{4}$
So, ${E}={\sigma}{A}{273}^{4}$------(1)
Now, for second one ${E} _{1}={\sigma}{A}{546}^{4}$-----(2)
On dividing equation(2) by (1), we get
$\dfrac{{E} _{1}}{E}=\dfrac{{546}^{4}}{{273}^{4}}={16}$

Or, we can say ${E} _{1}={16}{E}$

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

The rate of emission of a black body at temperature $27$$^{o}$C is $E _{1}$. If its temperature is increased to $327$$^{o}$C, the rate of emission of radiation is $E _{2}$. The relation between $E _{1} $ and $  E _{2}$ is :

  1. $E _{2}=24E _{1}$
  2. $E _{2}=16E _{1}$
  3. $E _{2}=8E _{1}$
  4. $E _{2}=4E _{1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know the relation for emissive power of body: $\dfrac { { E } _{ 1 } }{ { E } _{ 2 } } =\dfrac { { T } _{ 1 }^{ 4 } }{ { T } _{ 2 }^{ 4 } } $
${T} _{1}=27+273=300\ K$
${T} _{2}=327+273=600\ K$
So, putting all these data in above formulas, we get
${E} _{2}=16{E} _{1}$

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

The wave length corresponding to maximum intensity of radiation emitted by a star is $289.8$nm. The intensity of radiation for the star is :

(Stefans constant $=$ 5.6x10$^{-8}Wm^{-2}K^{-4}$, Wien's displacement constant = $2898 \times 10^{-6} mK$ )

  1. 5.67 x 10$^{8}Wm^{-2}$
  2. 5.67 x 10$^{4}Wm^{-2}$
  3. 10.67 x 10$^{7}Wm^{-2}$
  4. 10.67 x 10$^{4}Wm^{-2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given, Wavelength corresponding to maximum radiation $=289.8nm$
Stefans constant$=5.6\times 10^{-8}Wm^{-2}K^{-4}$
From Wien's law,
$\lambda _{max} T$= constant (b)
$b=2898\times 10^{-6}$
$\lambda _{max}=289.8$
$\therefore$ $T=\frac{b}{\lambda _{max}}=\frac{2898\times 10^{-6}}{289.8\times 10^{-9}}=10^4 K$

Intensity of radiation E
$E=\sigma T^4=5.6\times 10^{-8}\times 10^{16}= 5.6\times 10^8 Wm^{-2}$