Tag: observing space: telescopes

Questions Related to observing space: telescopes

Intensity of electromagnetic wave will be

physics observing space: telescopes maxwell's equations the nature of light introduction to electromagnetic waves

  1. $I=c{\mu} _{0}{B} _{0}^{2}/2$

  2. $I=c{\epsilon} _{0}{B} _{0}^{2}/2$

  3. $I={B} _{0}^{2}/c{\mu} _{0}$

  4. $I={E} _{0}^{2}/c{\epsilon} _{0}$

Show answer
Correct Option: B

The direction of propagation of electromagnetic wave is along.

physics observing space: telescopes maxwell's equations the nature of light introduction to electromagnetic waves

  1. Electric field vector, $\vec{E}$

  2. Magnetic field vector, $\vec{B}$

  3. $\vec{E}\cdot \vec{B}$

  4. $\vec{E}\times \vec{B}$

  5. $\vec{B}\times \vec{E}$

Show answer
Correct Option: D
Explanation:
The direction of propagation of the electromagnetic wave is always perpendicular to the plane in which $\vec{E}$  and  $\vec{B}$ lies.
So, the direction of the propagation of the wave, $\vec{C}=E\times B$.
So, the correct option is $(D)$

The electromagnetic radiations used for taking photographs of objects in dark.

physics observing space: telescopes maxwell's equations the nature of light introduction to electromagnetic waves

  1. $X$-rays

  2. Infra-red rays

  3. $\gamma$ rays

  4. $UV$ rays

Show answer
Correct Option: B
Explanation:

Answer is B.

Infrared (IR) is invisible radiant energy, electromagnetic radiation with longer wavelengths than those of visible light, extending from the nominal red edge of the visible spectrum at 700 nano-meters to 1 mm. 
Infrared is used in night vision equipment when there is insufficient visible light to see. Night vision devices operate through a process involving the conversion of visible light photons into electrons that are then amplified by a chemical and electrical process and then converted back into visible light. Infrared light sources can be used to enhance the available visible light for conversion by night vision devices, increasing in-the-dark visibility without actually using a visible light source.

Light appears to travel in a straight line, because.

physics observing space: telescopes maxwell's equations the nature of light introduction to electromagnetic waves

  1. Its wavelength is very small

  2. Its velocity is large

  3. It is not absorbed by surroundings

  4. It is reflected by surroundings

Show answer
Correct Option: A
Explanation:

Light appears to travel in a straight line because diffraction (or deviation from the path) is least in light. Diffraction is least because -of small wavelength of light. So small wave length of light causes the light to travel almost in straight line.

Which of the following rays have the highest frequency?

physics observing space: telescopes maxwell's equations the nature of light introduction to electromagnetic waves

  1. Radiowaves

  2. Infrared rays

  3. Gamma rays

  4. X-rays

Show answer
Correct Option: C
Explanation:

Gamma rays have the highest frequency.

The speed at which the light travels in vacuum is -

physics observing space: telescopes maxwell's equations the nature of light introduction to electromagnetic waves

  1. $\displaystyle 3\times { 10 }^{ 8 }m/s$

  2. $\displaystyle 3\times { 10 }^{ 3 }m/s$

  3. $\displaystyle 3\times { 10 }^{ 4 }m/s$

  4. $\displaystyle 3\times { 10 }^{ 10 }m/s$

Show answer
Correct Option: A
Explanation:

The speed at which the light travels in vacuum is $3×10^8m/s$

In vacuum, electromagnetic waves travel at the speed of

physics observing space: telescopes maxwell's equations the nature of light introduction to electromagnetic waves

  1. $3 \times {10}^{8} {m}/{s}$

  2. $3 \times {10}^{6} {m}/{s}$

  3. $3 \times {10}^{-8} {m}/{s}$

  4. $3 \times {10}^{18} {m}/{s}$

Show answer
Correct Option: A
Explanation:

In vacuum, electromagnetic waves travel at speed of $3\times { 10 }^{ 8 }$ m/s.

C = speed of EM waves
C = $\frac { 1 }{ \sqrt { { \mu  } _{ 0 }{ \varepsilon  } _{ 0 } }  } $
                ${ \mu  } _{ 0 }$ = permaebility of free space
                ${ \varepsilon  } _{ 0 }$ = permittivity of free space
putting the value of ${ \mu  } _{ 0 }$ and ${ \varepsilon  } _{ 0 }$ we get value of C approximately $3\times { 10 }^{ 8 }$ m/s.
                ${ \mu  } _{ 0 }$ = $1.257\times { 10 }^{ -6 }$ Henry/meter
                ${ \varepsilon  } _{ 0 }$ = $8.85\times { 10 }^{ -12 }$ Farad/meter

Reflection of a light wave at a fixed point results in a phase difference between incident and reflected wave of

physics observing space: telescopes maxwell's equations the nature of light introduction to electromagnetic waves

  1. $\dfrac{3\pi}{2}$

  2. $2 \pi $ rad

  3. $\pi$ rad

  4. $\dfrac{\pi}{2}$ rad

Show answer
Correct Option: C
Explanation:

Reflection of light wave at fixed point result in phase difference b/w incident and reflected ray of $\pi$. The wave gets inverted. Inversion behavior is noticed when medium is connected to dense medium.

Choose the correct answer from the alternatives given.


Which of the following has/have zero average value in a plane electromagnetic wave?

physics observing space: telescopes maxwell's equations the nature of light introduction to electromagnetic waves

  1. Both magnetic and electric fields

  2. electric field only

  3. Magnetic field only

  4. None of these

Show answer
Correct Option: A
Explanation:

A. Both magnetic and electric field.


Because both electric and magnetic field oscillate as sine and cosine wave, which over one period, average value is zero.

Which of the following statement is false for the properties of electromagnetic waves?

physics observing space: telescopes maxwell's equations the nature of light introduction to electromagnetic waves

  1. Both electric and magnetic field vectors attain the maxima and minima at same place and same time.

  2. The energy in electromagnetic wave is divided equally between electric and magnetic field vectors.

  3. Both electric and magnetic field vectors are parallel to each other and perpendicular to the direction of propagation of wave.

  4. These waves do not require any material medium for propagation.

Show answer
Correct Option: C
Explanation:

In electromagnetic waves,

1. The electric field and magnetic field varies continuously with time and have maxima and minima at same place and at same time.
2. Both electric and magnetic field have same energy.
3. both electric and magnetic field are perpendicular to each other and perpendicular to direction of propagation.
4. These waves don't require any material medium to propagate, they can propagate in vacuum as well.
So, the false statement will be the statement given in the option $(C)$
Hence, the correct option is $(C)$